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IN MEMORIAM 
 FLOR1AN CAJOR1 
 
<o 
 
 A 
 
 COLLECTION 
 
 DIOPHANTINE PROBLEMS 
 
 WITH 
 
 SOLUTIONS. 
 
 COMPILED BY 
 
 JAMES MATTESON. M. D., 
 
 DeKalb Centre, Illinois. 
 
 WASHINGTON: 
 ARTEMAS MARTIN. 
 
 1888. 
 
COLLECTION 
 
 OF 
 
 DI0PHA1VTINE PROBLEMS 
 
 WITH 
 
 SOLUTIONS. 
 
 COMPILED BY 
 
 J AMES MATTESON, M. D., 
 
 DeKalb Centre, Illinois. 
 
 WASHINGTON: 
 ARTEMAS MARTIN. 
 
 1888. 
 
CAJORI 
 
 Copyright, 1888, 
 
 By artemas martin. 
 
•'•' 
 
 n u*+ 
 
 M4a 
 
 INTRODUCTORY NOTE. 
 
 The following pages were left in sheets by the late Dr. Matteson at his 
 death, which occurred on Dec. 15, 1876. Had he lived, he would undoubtedly 
 have added many pages of interesting problems and solutions to the existing 
 number. 
 
 Having recently purchased from the guardian of the heirs the entire edition, 
 consisting of only a few hundred copies, I decided to issue it with the addition 
 of cover, title-page, and this note. 
 
 The authorship of the solutions, as near as can now be ascertained, is given 
 below. A considerable portion of this information has been furnished by Mr. 
 Reuben Davis, Sr., of Clifton, Kansas, one of the most ingenious of living 
 •'Diophantists," who rendered valuable assistance to Dr. Matteson in the 
 compilation of the work. 
 
 The solutions of Problems 1 and 4 are by the late Abijah McLean, Esq., 
 of New Lisbon, Ohio, who was very skillful in. handling Diophantine Problems 
 of great difficulty. 
 
 The solutions of Problems 5, 7, 9, 22 and 24 (except the last two paragraphs ' 
 of the solution of 22 which are by Mr. Davis) are from the pen of the late 
 lamented Dr. David Sherman Haet, of Stonington, Conn., who was one of 
 the ablest, if not the ablest, of the Diophantine writers of his time. 
 
 Dr. Hart also rendered efficient aid to the compiler in the revision of many 
 difficult solutions. 
 
 The solutions of Problems 6, 8 and 12 (except the last two paragraphs of 
 the solution of 12 which are by Mr. Davis) are by Dr. Matteson. 
 
 The other solutions are mostly the work of Mr. Davis, with some slight 
 modifications by Dr. Matteson. 
 
 The following errors have been noted by Dr. Hart, Mr. Davis, and the 
 writer : 
 
 Page 4, line 3 from the bottom, for "because y is odd" read because x is odd. 
 
 Page 7, line 23 from bottom, for "three squares sought" read three equi- 
 different numbers from which are obtained the three squares sought. 
 
 Page 8, line I, for "of the natural series" read in the natural series. 
 
 Page 8, line 2 from bottom, for "nad" read and. 
 
 Page 9, where ' ' of the natural series" occurs, read in the natural series. 
 
 Page 11, line 13, for "13 = 32 + 12" read 13 = 32 + 22. 
 
 Page 14, formula (14) should be 
 
 _ [( - ab + ac + 6c) 2 - 4a6c 2 ] 2 
 
 ~ 8abc( -ab + ac + bc)(ab -ac + bc)(ab + ac- be) ' 
 and to preserve the sign of x unchanged in (14) as well as in (15), the relations 
 between a, 6, c, must be such that the sum of any two exceeds the other. 
 
 Page 15, line 3, right-hand member of equation, for " - 6 2 c 2 " read - 36 2 c 2 . 
 
 ARTEMAS MARTIN. 
 Washington, D. C, July 2, 1888. 
 
 A 
 
Digitized by the Internet Archive 
 
 in 2008 with funding from 
 
 Microsoft Corporation 
 
 http://www.archive.org/details/collectionofdiopOOmattrich 
 
Diophantine Problem. 
 
 It is required to find four affirmative integer numbers, such that 
 the sum of every two of them shall be a cube. 
 
 Solution . 
 
 If we assume the first=J(x 3 +^ 8 — s* ), the second— ^(z 3 — if-\-£ ), 
 the third=J(— tf+tf+s?), and the fourth = v 3 — £(.r 3 -f y 3 — z 3 ) ; then, 
 the first added to the second^x 3 , the first added to the thirdly 3 , 
 the second added to thirds* 3 , and the first added to the fourth=t> 8 . 
 Thus four of the six required conditions are satisfied in the no- 
 tation. It remains, then, to make the second plus the fourth= 
 v 3 — y SJ r z 3 =c\\be, say— «? 3 , and the third plus the fourth='y 3 — x 3 -]- 
 2 s =cube, say=w 8 . Transposing, we have to resolve the equalities 
 ^J r ^ == ic s ^-)fz=u 3 -{-j^ ; and with values of x, 3/, 2, in such ratio, 
 tli.it each two shall be greater than the third. 
 
 Let us first resolve, in general terms, the equality v 3 -\-z*~w s -\-y 3 . 
 Taking v=a-\-b, *=a — b, w=c-\-d, y=e — d, the equation, after 
 dividing by 2, becomes a(a 2 -\-3b 2 )=c(c 2 -\-3d 2 ~). Now assume 
 a=3np-\-3/nq, b=mp — 3nq, c=3nr-{-3ms, and d=mr — 3ns. Sub- 
 stituting these in the preceding equation, it becomes (3np-\-3mq) 
 [(3np+3mqy+3(mp — 3ra£) 2 ]=(3rcr+3ms)X[(3rar+3ra*) 2 +3(mr— 
 3nsy] ; or (3np+3)?iqXp 2 +3q 2 )3(m 2 -\-3ri 2 )=(3nr+3ms)(r 2 +3s 2 )X 
 3(m 2 -\-3?i 2 ) ; which, dividing by the common factors 3.3(m 2J r 3n?), < 
 reduces to (?i2? J r mq)(p 2J r 3q 2 )=(?ir-{-ms)(r 2 ^ r 3s 2 )' y or, np(p 2 -\-3q 2 ) 
 +mq(2> 2 +3q 2 ')=nr(r 2 +3s 2 )-{-ms(r 2 -\-3s 2 ); 
 
 . • . m : n : : r(r 2 +3s 2 )— -p(p*+H*) ' q(p 2 +3q 2 )—s(r 2 +3s 2 ) ; and, if 
 we take m=r(r 2 +3s 2 )— p(p 2 -\- 3q 2 ), then n=q(p 2 +3q 2 )— «(r 2 + 3s 2 ). 
 a=3np+3mq=(3rq— 3ps)(r 2 +3s 2 ), 
 b=mp—3?iq=(pr-^3qsXr 2 ~\'3s 2 )—(p 2 -^3q 2 )\ 
 c=3nr+3ms=(3rq—3ps)(p 2 +3q 2 ), and 
 d=mr— Sm=(r*+S8*)*— (pr+SqsXp^+Sq 2 ), and, at once 
 v=a+b=(Srq— 3ps+pr-{-3qsXr 2 +3s 2 )— (jt> 2 +3^ 2 ) 2 , 
 z=a—b=(3rq—3ps—pr—3qsXr 2 -\-3s 2 ~)+(p 2 -)-3q 2 ) 2 f 
 w=c+d=(3rq—3ps—pr—3qs)(p 2 +3q 2 )+(r 2 -\-3s 2 y, 
 y=c—d=(3rq—3ps-\-pr+3qs)(p 2 -{-3q 2 )—(r 2 -{-3s 2 y. 
 We have thus arrived at general expressions for the values of v, z r 
 w, y, such that v*+s?=w*+y i =18(rq— ps)(r 2 -f-3s 2 )(j» 2 -f 3q 2 )X 
 [( r 2 +8s2 )3_ ( 2 i? , r -|-6^)(r 2 +36- 2 )(^ 2 +3^ 2 )+( i > 2 +3^) 3 ], as will ap- 
 pear from actual involution, &c. 
 
 But, to have the final answer in positive numbers, v must nearly 
 =32, and w nearly=3;y. 
 
 Taking p=6, q=U, r=1, and «=14 ; there results 
 
 ^=48.49.14.13— 48 2 .13 2 ==48.49.14— -48 2 .13==13.2976, 
 z=— 49.49.12.1 3+48 2 .13 2 =— 49.49.12+48 2 .13=13.1140,, 
 w=— 49.48.12.13+49 2 .13 2 =— 49.48.12+49 2 .13=13.2989, 
 2/=48.48.14.13— 49 2 .13 2 =48.48.14— 49 2 .13=13.1043. 
 Having now, (dividing by the common factor 13), 
 (2976) 3 -|-(ll40) 3 =(2989) 3 +(l043) 3 =278387l4l76=7 3 .3 3 .4 3 .13.361^: 
 
2 
 
 it remains to find two other cubes, u* and a 3 , such that w 3 -}-^ 3 — 7 3 .3 3 . 
 4M3.3613. This we shall accomplish, if we succeed in dividing 
 l3.3613=(l*+32 2 ).(55 2 +3.l4 2 ) into two cubes. Now, it is demon- 
 strable that when the sum of two cubes is of the form (a 2 -\-3b 2 ){ 
 c 2 -\-3d?) ; and the relation of the quantities is such that (a 2 -f3# 2 )= 
 either 3d — c, or c — 3d, the cubes themselves will be 
 !0 2 +3& 2 -j-c-M) 8 and £(> 2 +3£ 2 — c— d)\ for the sum of these cubes 
 equated to the assumed sum, is 
 
 or (a 2 +3£ 2 ) 2 +3(c+d) 2 =4(c 2 +3^), 
 
 or (a 2 -^3b 2 y=c t — 6cd+9d?=(c— 3d) 2 =(3d— c) 2 . 
 
 Wherefore, a*-\-3b 2 =c— 3d, or, —3d — c, as was to be shown. 
 
 In the example before us, #±±1, b—2, c=55, d—14, anda 2 -j-3& 2 = 
 
 13=c— 3d=55— 3.14; then -|(a 2 +36 2 -f-c-j-d)==41, and l(a 2 +3b 2 — 
 
 c— d)=— 28, and (41) 3 — (28) 3 =13.3613; and, instead of the sum, 
 
 we have found the difference of two cubes=13.3613. 
 
 But, Mathematicians have shown that 
 
 x a (o8_ 26»)\» Sb(2a?— b 3 )\ 3 . ,., 
 a 3_fls = ^____j +^ a »-j-y ^ ; m which, taking a=41, 
 
 #=28, we immediately obtain 
 
 (1081640V , /341899V \ 
 -^29T>)+V-3029r)= 13 ' 3613 - 
 
 Multiplying, now, by the reserved cube factors, 7 3 .3 3 .4 3 , we find 
 
 30285920 L 9573172 . ,,.,., 
 
 u== . , and x= i QO q 7 i values that likewise have the prop- 
 
 er ratio to insure positive numbers in the question under solution. 
 Thus, for v 3J r z 3 =w 3J r y 3 =u 3J r x 3 , we have found 
 
 or, since y is odd, multiply by 10097X2, for integers, and we have 
 ^=60571840, :r=23021160, y=2l062342, 2=19146344. 
 With these values, we at once obtain the following answer: 
 £0*?-rY— z 3 )=20809l3082956455l42636, 
 ^(a?— 2/ »+2 8 )=493780l347510680732948, 
 ^(_05»+i/ 8 +& 3 )=726281O47641O016163O52, 
 v z — i(^+2/ 3 — 2 3 )=214972108693241589340948. 
 Being four numbers as required, such, that the sum of every two 
 of them is a cube number; resulting in the six cubes, (19146344) 3 , 
 (21062342) 3 , (23021160) 3 , (60097344) 3 , (60359866) 3 , and (60571840) 3 . 
 This answer, it will be observed, is the same as that contained 
 in the Mathematical Miscellany, which was communicated by 
 Wm. Lenhart, where it was found from an inspection of a Table 
 of Cube Numbers. Here, the given solution is independent of 
 Tables, and we believe the numbers found to be the least possible. 
 OBSjERV*jMTIOJY : With regard to the assumed numeral 
 values of p, q, r, s, in this solution, we began with the smallest 
 integers, 0, inclusive, and soon found the numbers 6, 14, 7, 14, to 
 answer the purpose in hand, namely: to produce four cubes such, 
 
3 
 
 that v*-{-2?=w*-{-y*, and which shall have the proper ratio to pro- 
 duce positive numbers in the answer. There are, however, an in- 
 definite number of values for p, q^ r, 6, that will result in precisely 
 the same values of v, z, to, y, when reduced to the lowest terms. 
 
 In fact, any values of p, q, r, «, of the modulus 
 p=:mp f -]-Sn(f ; q= — np'-\-mq f ; r=mr'-\-3ms' ; s= — wr'+ras'; will, 
 after dividing, result in the same values of v, z, w, y, since m and 
 n thus disappear, and the general expressions for v t *, w, y, remain 
 of precisely the former modulus. Thus, if p\ q\ r\ s\ be respec- 
 tively 6, 14, 7, 14, and m=l, n=2 ; thenp=90, q=2, r=91, s=0, 
 and the same answer results, after dividing by (m 2 -{-3tt 2 ) 2 =169. If 
 p'=6, £'==14, /=7, s'=14, m=l4, n=9 ; then p=462, ^=142, jt= 
 476,s=133 ; and now, if m'=55, n'=l4, then jij>=31374, ^=1342, 
 r=3l766, s=651 ; all which will result in the same answer. 
 
 We have given a successful method of finding two cubes, whose 
 s^m=13.3613; or rather, in the first instance whose difference^ 
 13.3613; and here we desire to add, that when such cubes are In- 
 tegral, they may be found by assuming/-^, and/ — g for the roots, 
 when the sum is an even number ; or \{f-\-g) and %(f — </), when 
 the sum is odd In this case it is odd ; then ^f-\-gf-\-^{f — 00*== 
 13.3613; consequently, /(/ 2 -j-3<? 2 )=4.13.3613. 
 
 Now, since 3613 is a prime number, evidently greater than /', 
 f, if an integer, must be one of the factors, 1, 2, 4, 13, 20, or 52. 
 On trial, we succeed only with 13=/', in which case, g—Q9^ and 
 thence -J-( # /*-[-<7)=^41, and ■£■( f — g)= — 28, as before found. 
 
 •In investigation far a different answer. 
 
 We have, on the first page, 
 
 a=(3rq— 3/w0(r 2 +3* 2 ), 
 
 b=( pr+ 3qs~)(r 2 + 3s 2 )— (> 2 -f 3<f )*, 
 
 e=(3rq— 3^5)(j? 2 +3^ 2 ), and 
 
 d=^J r ^s 2 y—(pr-i-dqs)(p s -\-Sq 2 ). 
 Let us now make r 2 -\-3x 2 divisible by p' 2 -\-3q 2 . This is done by put- 
 ting r=3n f q-\-m'2ii and s=m'q — n'p ; for then r 2 -\-3s 2 becomes 
 (m' 2 -\-3?i' 2 )( p 2 +3q 2 ). By substitution, we obtain 
 
 a=3n'(m" i +3n' 2 )(p 2 +3q 2 )\ 
 
 b=m'(m' 2 +3n' 2 ) ( p 2 ^-3q*) 2 — ( p'+Sq*)*, 
 
 c=3n f (p 2 -i r 3q% and 
 
 d=(m' 2 +3n' 2 ) 2 ( 2 ? 2 +3q 2 ) 2 —m'( p 2 +3q 2 )\ 
 Dividing now by the common factor (p 2 -\-3q z )\ we have 
 
 a=3ri(m' 2 +3ri 2 )=Sgh(f t -\-3g 2 ) J 
 
 b=--m'( m' 2 -\-3n' 2 } — 1 =fh( f 2 -\- 3g 2 ) —h\ 
 
 c=$n'= Sg h\ and 
 
 d=(m*+ Sri 2 )*— m '=( t P+3g 2 )— fh* ; that is, after putting 
 m r =f-i-h, n'=g-i-h) and multiplying by A 4 . 
 
 Whence, (and changing signs for positive numbers) we have the 
 following theorems : 
 
 a+b=v=K3g-f)(f 2 +3g 2 )+h\ 
 
 a-b=z=h(3g+f)(f 2 + 3g 2 )-V i 
 
Beginning with the least numbers, we soon find that with /=5, 
 <7=8, and A=14, the least values of t?, fc, w, #, to answer the ques- 
 tion, will result. Then v, », i£, //, will be, respectively, 90138, 
 49686, 99225, 32487 ; which, by dividing by the common factor 
 147=(0 2 -|-3.V 2 ), become 654, 338, 675, 221 ; so that (654) 8 -j-(338) 8 
 =(675) s ~f-(221) s . These are smaller numbers than those before 
 found,, within the relation to produce affirmative results. But, in 
 attempting to find two different numbers, u and .r, such that i^-j-x 3 
 —the same sum, by either of the methods herein before employed 
 successfully, we fail. We are, therefore, in this instance, driven 
 to the following prolix, but equally well-known method : 
 
 It. is known that rf+P={ \p__p ) —\ >__l^ ) ' in 
 
 which, substituting «=r675, and #=221, we obtain 
 
 (675V L (22 ^^( 222165852975 V_( 138321162031 V_ 
 (bto)-\-{Z*L)—y 296753014 / V 296753014 / ~ U * * 
 
 and from another equally well-known Theorem, we next have 
 
 u ZJ r x z =\ 
 
 222165852975 138321162031 
 
 b = <>Qfi7zzmA ' Wherefore, 
 
 296753014 7 296753014 
 
 b'(2a' 3 —b' z ) 2667483866296141146514629941994412921894289729 
 a'^b'* ~~~ 4039417275545528891152791084460320436090324 ' 
 
 X=z 
 
 a'\a'*—2h z ) 1260271634077571065592737296883736641712234175. 
 tiF+F* ~~ 4039417275545528891152791084460320436090324 
 We have now the numeral values of if, % ?r, r, y, », such that 
 t7 8 -|-« 5 =w? 8 +^ 8 ='W 8 +-t ,s ; and, at the same time, those of :r, y, e, are 
 such as to insure all the numbers, in answer to the proposed ques- 
 tion, to be affirmative. Multiplying all the numbers by the com- 
 mon denominator, in the above fractional values of u and x, we get 
 integral values, as follows: 
 
 M=2667483866296141 14651 562994199441 2921894289729, 
 v=264l7788982O6775894813925369237O4956520307l896, 
 «?=2726606660993232001528133982010716294460968700, 
 ^•=1260271034077571065592737296883736641712234175, 
 #=892711217895561884944766829665730816375961604, 
 z=1365323039134388765209643386547588307398529512. 
 These numbers are believed to be prime to each other, and are, 
 therefore, irreducible by any common factor; and, if substituted in 
 the notation with which we set out, namely: 
 
 H'^.'/ 3 -* 3 ), i(^-'/ 3 +^), *(-*•+/+*•), * , -K*-H*-*fc 
 
 or (because y is odd) in the equivalent expressions 
 
 *(**+&*— *U H r "— V 3 -f* 3 ), 4(— .* 3 -f*/ 3 +* 3 ), Sv 3 — 4(jL 3 -\-f— 2 s ), 
 will give, four numbers such as are requiredan.this. problem. 
 
2. Find three integral numbers in arithmetical progression, 
 such that their common difference shall be a cube; the sum of any 
 two, diminished by the third, a square; the sum of the roots of the 
 required squares an 8th power; the first of the required squares 
 a 7th power, the second a 5th power, the third a biquadrate, and 
 the mean of the three required numbers a square. 
 
 Solution. Let [(*- 2 -.t-|-1)tA (f+l)y*, and (V-L-Z+ !)</*] , . . [a ] ? 
 represent the three numbers in arithmetical progression, their com- 
 mon difference being xy*. Then must «{ (x 2 — 2x-\-V)y 2 z=.[^\ ... [1], 
 (s*+i)2,'=n . . . [2$ and (rf+2r+ \)}f=U • ■ • M J> • • • [A]. 
 
 [I] and [3] are squares. To make [2] a square, let a?-J-l=(a: — p~) 2 \ 
 then will x=z(p 2 — l)-?-2j). This value of x in [vl] changes it to 
 
 ^ V-^r— ; ?/ 2 • • • M, (^ J » • ■ • M. V — ^ — J * 
 
 ... [6] ^ • • • [7?]> all squares; the sum of their roots being 
 
 (3r>2 i\ 
 ~2~ — ) :V • • • [?]• Let y=pz 2 , ^ ien [-#] atl d [7] become 
 
 * (tt3l)V.. . [8]) (^Jv.:: . [•], (£±*=±)V 
 
 . . . [10], and (^2 == ^)^ • • • [H] ^ • %[<?]. [H] is a square and 
 [10] a biquadrate when ^2=1. Putting ^-[-1 for j», [10] and 
 
 [II] will be (§H-2?+ 1)V . . . [12], and ( 3 |^-3£+l)V . . . [13]. 
 
 <7 2 4'r-}-l) 
 
 Assume — -\-2q-\-\=(qr — l) 2 , then Q==~n~rzi ' Substituting 
 
 24(r-\-l)' 2 12 (v-\- V) 
 this value of y in [13], we obtain ^_ l i + 2 r 2 —l + 1 • 
 
 Adding these terms, and rejecting the square denominator, the 
 
 result is 4r 4 -f24r 5 -f 44r 2 -j-36r-}-13, which make==(2r s +6r-f 2)?; 
 
 3 4(r4-l) 
 then will r= — -j-, ff— ~ 2 =8, andjtteg-f-l^O. 
 
 Substituting this value of p in [ (7], we shall have 
 «{ 81*** . . . [14], 41V . . . [15], W • • • [16], U*# . . . [17] y,,i II)-]. 
 Take 2=llV, and [7>] becomes <( 31 2 .ll 12 i> 16 . . . [18], 41 2 .11 12 « 16 . . . 
 [19], 7 4 .11 1 V 6 . . . [20], UV . . . [21] y. . .,[J£], and the common 
 difference, &2/ 2 =360.11 12 -y 16 . . . [22]. To make [22] a cube, let «=: 
 5 2 .3w 3 ; then [j£] and [22] change to <( 31 2 .11 12 .5 32 .3 16 m> 48 . . . [23], 41 2 . 
 }l n .5**.3"w« . . . [24], 7 4 .ll 12 .5 32 .3^o 48 . . . [25], 11 8 .5 16 .3V 4 . . . [26], 
 1 l 12 .5 33 .3 18 .2 3 ^. . . [27] j*. . . [i^]. [23] is a 7th power when «e= 
 81».11 6 .5 4 .3W, which changes [.F] to «{ Sl 98 .!! 252 ^ 224 ^ 112 ^ 336 . . . [28], 
 41 2 .31 9 ^11 252 .5 224 .3 U V 536 . . . [29], 7 4 .31 96 .11 252 .5 224 .3 11 V 36 . . . [30], 31 48 .11 128 . 
 5 H2 3 56 w i68 . f t j- 31 ] ? si 96 .!! 252 ^ 225 ^ 114 ^ 336 . . . [32] y . . . [6?]. 
 
 To make [29] a 5th power, put u=41\3l\ll\5.3H 5 , and the 
 expressions in [#] become, respectively, 
 
 (41 504 .31 721 .1 1 630 ^ 280 ^ 560 ^ 840 ) 2 ^^! 144 ^ I 206 .! 1 180 .5 80 .3 160 ^ 240 V 
 (41 505 .31 720 .11 630 .5 280 .3 360 ^ 840 ) 2 =(41 202 .31 288 .11 252 S 112 ^ 224 ^ 336 ) 5 
 (4l 504 .31 720 .l l 630 .7 2 .5 280 .3 560 ^ 840 ) 2 =:(4l 252 .3i 360 .i l 315 .7.5 140 3 280 t i20 Y 
 

 41 504 .31 720 .11 632 .5 280 .3 5C0 ^ 840 =(41 63 .31 90 .11 79 .5 35 .3 7 ¥ 05 ) 8 , 
 
 4-1 1008 g^HlO JJ1260 g561 gll22 23^1680__/^^336 g-j^480 j|420 ^187 33"! # 2^ 560 ) 
 
 In the last five lines above, eight of the nine required powers are 
 found; it remains to determine the required numbers; the mean of 
 which will be the ninth of the required powers. 
 
 Since x=(p 2 — l)-4-2/?,/>=9, and y=pz 2 , the expressions in [«] 
 are 1321* 4 , I68I2 4 , and 2041* 4 . But ss=HV changes them to 
 (132l.ll 12 .v 16 , 1681.11 1S V 6 , 2041. ll 12 -y 16 ) .. . [//] ; v=5 2 .3w* makes 
 these 1321.11 12 .5 s2 .3 16 w i8 , 1681.1 1 12 ^ 32 ^ 16 ^ 48 , and 2041.1 l 12 .5 32 .3 lf V s .; 
 and ?0=31 2 .11 6 .3 2 .5 4 m 7 , transforms these last to 
 1321.31 % .ll 252 .5 m .3 112 « 336 , 
 
 lesi.si^ii 252 ^ 22 ^ 112 ^ 336 , 
 
 2041.31 96 .11 252 .5 224 .3 112 ^ 336 . 
 Finally, w=41 3 .31*.ll s .5.3 3 J 5 ; hence, the numbers sought are 
 
 ^iwsiwii^.swa^^^iesi^i'.si^^ii'.s^^^ 
 
 204I.41^^lHll».5«.3^^ M *^2041(41^81 1 ^.(llU^8^ u ) 1 ^^ 
 
 the numbers being simplest when t=l. 
 
 Note. [10] may be made a 4th power and [11] a square by other 
 
 methods. If the quantities within the parentheses be multiplied 
 by 4, the result is 2p 2 -^~4p—.2=\J t . . . [a], and 6p 2 —2=[J . . . [b]. 
 Let q-\-l=2); then [a] and [b] become 
 
 2tf+&q+*= Q =b' 2 - • • [»1 »»d 6<y 2 -|-l2(7-|-4=rn^:« 2 . . . [b']. 
 Subtracting [a'] from [b'], and factoring the difference, we obtain 
 
 <j(4q-\-4)=(a—bXa -|-A). Take q=a—b, and 4£+4=a+& 
 Adding these two equations we have a=(oq +4)-s-2. Substituting 
 this value of a in (b'), and reducing, we find </— 8, and thence p=8. 
 
 To find a third value of p, put 8-\-9=p'; then the quantities in 
 the parentheses of [10] and [11] become 
 
 •2*' 2 -H0H-169=n=^' 2 • * ' M' an(1 6s 2 -f-108s+484=n=«,' 2 . . . [ c \. 
 Multiplying [d] by ll 2 , and [c] by 7 2 , to make the numeral squares 
 169 and 484 equal, Ave obtain 
 
 2426- 2 +4840s-j-237l6=rf 2 . . . [d'], 294« 2 -f.5292.<}-f237] Q=c 2 . . . [c'j 
 Subtracting [d'] from [c'], and factoring the difference, we have 
 *(52*+452)=(c— d)(t'-+ r d). Put s=e— d, and 52*4-452=^+^ ; 
 then, by subtraction, d=(5ls-\-45'2)~2. Substituting this value 
 of d in [d'], and reducing, Ave find j»/= -J-J-H • 
 
 3. It is required to find three whole numbers in arithmetical pro- 
 gression, suchthat their common difference shall be a cube; the 
 sum of any two, diminished by the third, a square; and the sum of 
 the roots of these squares a square. 
 
 Solution. Let x 2 —xy-^y\ x 2 +y\ and x 2 +xy-\-y 2 
 represent the numbers in arithmetical progression, whose com- 
 mon difference is xy ; then must 
 x 2 —2xy+y 2 =[J . . . [1], x 2 +y 2 =C] . . . [2], and &+2xy+y>=Q . . . [8} 
 
 [1] and [3] are already squares, and [2] is a square when 
 x=r 2 — s 2 and ?/— 2rs, y being taken > sc. Whence, by substitu- 
 tion, [1], [2], and [3], are changed to 
 
(— r'+Srs-M 2 ) 2 ^ 2 -hs 2 ) 2 , ilit^+ar*-^)"-; 
 the sum of whose roots, r 2 -j-4r#+ s 2 • • • [4], must be a square. 
 
 Put r*+4rs+s*==(r4-ns?i then we find r=^zQL 
 
 ^ . 3 ,. 5s 9s 2 10s 2 , 906'* 3 2 .10s* 
 
 lake n=—', then r= -j-, «==^> y=±qp, ««^ ajy=a~jf=e=--y- "= 
 
 a cube . . . [5] ; and the assumed numbers become, respectively, 
 1321s 4 r,.-! 1681s 4 r „, a 2041s 4 rol 
 
 -^—- • • M, — gp- - • • PI *d& — iF — • • [«]• 
 
 But [5] must be a cube, which is the case when #=3.1 W 1 , for then 
 
 *#= — ^ — = ( — S~ J == a c,11 ' je ' an< * L 6 J' ['1 an * L 8 1 become 
 132l.5 8 .3W-=4l797265625?' 12 , 
 1681.5 8 ,3 4 tf 12 =5318'7890625£ 12 , 
 
 3041.5 8 ,3¥ l2 =645785156.25* 12 ; where t may be any 
 integer — the least numbers being when t~l. 
 
 If s-— .2 :i .3.10 2 £ 12 , or £=2 12 i; 12 , the above numbers become 
 1321.10 8 .6V 2 ^171201600000000*> 12 , 
 lG8l.l0 8 .0^' 12 ^21785760000OOO0r 12 , 
 204lJl0 8 .6V-=26451^600000000y 1 l 
 Substituting the values of r, x, and //, in terms of s, and taking 
 «=11 8 .2V, expressions [1], [2], [3], [4], [5], become [18] to [22], 
 inclusive, and [6], [7], and [8], become the three expressions m 
 [//], in the solution of problem 2. 
 
 Again, take 1321.5 8 .3^ 12 , 1081.5 8 .8 4 ^, and 2041. 5 8 .3 4 * 12 , for the 
 three squares sought, in the second condition of problem 2. 
 Tken must 31 2 .5 8 .3 4 * 12 ^ a 7th power, . . (1), 
 U 8 .5 8 .3¥ 2 = a 5th power , . . (2), 
 48 a .5 8 ;* 4 £ la 3= a 4th power. . . (3), and the 
 Bum of the roots of these squares, ll a .5 4 r3W=an 8th power. . . (4). 
 (3) is already a 4th power, and expunging the 7th power factors 
 from (1), the 5th power factors from (2), and -extracting tbe-squar 
 root of (4), we have only to make 31-.5.37 r '- a 7th power . . . (5), 
 41 2 .5 3 .3 4 fc-- a 5th power . . . (o), and 11.5*.3^= a biquadrate . . . (7>). 
 (7) is solved by taking fcll.5 2 .3?e\ Whence, by substitution^ 
 (5) becomes $V.ll*Jf a .&u*= a 7th power, and (6) becomes 
 41 2 .ir-.5 7 3V— a 5th power, or, expunging the 7th and 5th power 
 factors, as before, 31 2 .11 5 «5 4 .3V= a 7th power . . . (8)., and 
 il».ll*.5 a .3w 8 = a 5th power . .. . (9). Take w=^31 2 .l 1 5 .5 4 .3V, and (8 | 
 is satisfied. .(9) then becomes £1*81*.] l n .5 u .3V' 21 ^ a 5th power, or, 
 cancelling 5th power factors 41*.31.11*i5 4 .3*l>3= a 5th power. . . (10).. 
 JJence t*&=41'.31'.ll 1 .3r\0 B solves (10,). 
 
 Retracing, we have ^41 2 ^1 28 .11 2 ^5\3 2 V :;5 , 
 
 . • . «==41 fl .tri» ] 1 *5 u .S»tt>* u 4 . 41 s *;31 120 .l l***&Hf% 
 m - . t=41 u .'3l m .l l 105 .5 4C .3 a V U0 , 
 
 ^12 ^1008 g T H40 j j 12(10 *6SI ^UKi^.1680 
 
 £ 8 i S 4 $ 1 *=41 M88 .31 MW . 1 1 lM ,S 9n .d h ^w im ==m i 
 
 - •. 1321;;?, 108b/?, and .2041 m 9 sat isfv .all the condition*. 
 
8 
 
 4. To find n consecutive numbers of the natural series, such 
 that the sum of their cubes shall itself be a cube, n being= a cube. 
 
 Solution. Let x represent the first number in the series, then 
 must x^}-(u:-|-l) 3 +(H-2) 3 +(^+3) 3 -j-&e. to (x+n— l) s =cube. 
 To obtain a more manageable expression for the sum of this series, 
 
 wo. know that l°-|-*°+3'+,fcc. to *= «+**+»' = (**££) ' ; 
 
 !( nd, .-. , that l 3 +2 ! +3'+&c. to (g— lj»=r A g ~ 1)x j ; and- 
 
 hc-nce, that l»+2'+3»+&c. to (*+«-!)»= (^±tZ^4^)* 
 
 Wherefore, the sum of all the cubes of the required series of wNos. 
 
 will be represented by ( g J — I 9 — J ==0*, cube. 
 
 . • . (x+n— l) 2 (-r-f nf— (x— l) 2 x 2 r=4a 3 . 
 But the difference of two squares is equal to the product of the 
 sum and difference of the roots ; therefore, 
 
 [fx-\-n — y){x-\-n)-{-(x — l)x~\.[(x-\-n — l)(.r -[-??,) — (x — l)x]=4« 3 . 
 Or, [2^4-2(w— l)a_j-(M— i^ n ](2r+n— l)n=4a*. 
 Here, n is, evidently, a factor of « :J ; and since n is a cube number, 
 ■t'% t is a factor of a; and we may represent 4a 3 by 4W, and 
 substituting f for ft, and dividing the equation by t 3 , it becomes 
 
 "[2c»+2(^— l^-K* 8 — l)^ 3 ](2r+^ 3 — 1)=4^ 3 . 
 Making the multiplication in the left-hand member, and also mul- 
 tiplying both sides by 2, the equation is 
 
 s, :; -l2(£ 3 — l)x 2 -u4(? 3 ~l)(2^ 3 — l).r-f2(7 3 — iyP=$P=i?bub&. 
 Now tli i s being one of those cubic formulas, irreducible by means 
 of an assumption that will destroy two tefrms, we must avail our- 
 selves of the next expedient, viz., the vanishing only of the term 
 s;< 3 ; and, in accomplishing this, in order that the utmost advant- 
 age of a division by factors may be had, that is, that the resulting 
 equation may be divisible by (t — 1)(H -1), let us assume 
 2£=2r-HbH-l) 8 -l- 1]0- 1); then will 
 12(^H-2^2)(^-l)/^-(i' 2 4-2^2) 2 .6(?-l) 2 x+(^+2H2) 3 (^-l) 3 ^ 
 s// ! . Equating these two equal cubic expressions, rejecting the 
 common term Sr 1 , and transposing, we have 
 12[(^| 2^4-2)C?-l)-(^ 3 -l)]x 2 +2[(^H-2'-f2) 2 .3(?-l) 2 -(i 3 -l).2(2^ 3 -l)> 
 2(f ..- iyp— (t>~\-2t+2y(t— i) 3 . 
 
 Or, 12^— l> 1 +2[0 s +2«+2) 2 .3C#— 1 ) 2 — (f— l)XW— l)]-r= 
 2if—lYP— (?+2t-\-2)*(t— 1)*. Obviously, this equation is divisi- 
 ble by £+1, and also by t — 1, and it reduces to 
 
 I2r 2 — 2(^—6^— 2« 2 -|-10).r— # 7 — 3^— 2^— 2^-r-10i 8 4-4« 2 — 8. 
 Multiplying by 12, and adding (t 4 — 6f— 2£ 2 -fl0) 2 to both sides, 
 the quantity on the left-hand side will necessarily be a complete 
 square; and the result will be 
 
 [12a— (t*—GP— 2£ 2 -fl0)] 2 ^ 8 — 4*°-f 8* 2 -j-4=0 4 — 2* 2 — 2) 2 , 
 12a-— (<*— 6f— 2* 2 -f 10)=«*— 2* 2 — 2, nad 
 :«. 4(tf*- 3f 3 — 2^-^-4); Where £ 3 - =», or n must be a cube number. 
 
 .. 
 
6, To find a cube number oi numbers which are cubes whose 
 roots are consecutive numbers of the natural series. 
 
 Solution. Let (.r-[n-l]) s , (.r_[> -2]) 3 , (x-\_n-Zj)* * 
 
 . . . . (^ ; _j_[ n _3])3 5 ( a ._^_[^ l _2]) 3 , (, r _L[y A — ]]/•, be an odd series of 
 cube numbers whose roots are consecutive numbers of the natural 
 scries; r 3 being the middle term. 
 
 Then, beginning at this term, the sum of one term is ./ :; . 
 Ab«», the sum of three terms is (.r— l) 3 +./ 3 -f (.r-{-l) 3 = 3-* 3 -f6.r; sum 
 of five terms is (,-LO :5 -i-(.r-l) :5 +^+{x-+l) 3 H-(^-r^) 3 =-- i*+*<to 
 In like manner, the sum of seven terms us = 7./- 3 -r84.^ 
 
 ami the sum of nine terms is — Qx?~\~l$Qx 
 
 ' ■ ' eleven terms is = lltf+ZiOx 
 
 1 ' ' thirteen ' ' = 13.> 3 -j-546r, &c. 
 
 . • . the sum of 2jt—l terms is=(2y^ — l)->. s -\- (2y/ 3 — :W-{-n).r . . . [1], 
 lor the nth term of the series 1, 3, 5, 7, 9, 11, 13, «fec, is %n — 1, 
 and the («— l)th term of the series 6, 30, 84, 180, 330, 546, &c, is 
 hr? — 3u' 2 -{-n. But the problem requires that [1] shall be a cube, and 
 also that 2^-1 shall be a cube. Let 2n-l—p* ; then »=£(/>*+ *)j 
 and by substitution, formula [1] becomes 
 
 \>s/ p9 i \ r/ 5 — 1 
 
 v .r = a cube; or, dividing byy/ ! , '"+-7 — 8?^= a cube- 
 
 This expression will be a cube when «=4« Let ;G=^4Hb then by 
 substitution, multiplying by 8, and arranging the terms, we have 
 !2 7 - : (2y> 6 - L 4)y- r yA^a cube, which put==(2#-f-/» s ) 8 . Whence, 
 ;>«— V+2 (p 2 -l) 2 -3 , Q— *)'_.. (i>— 1)'Q + 1)' 
 
 /7 " 6(p 2 — 1) — • J ~ 6 
 
 Here the value of* must be integral, to have consecutive numbers. 
 This will be effected by taking p= 6/>/zbl ; m being any number. 
 
 Let m=l; then, using the negative sign, j0=5, .*— 96, and y/~63. 
 Substituting these values in the original series, Ave have 34, 35, 36, 
 .... 156, 157, 158, for the roots of 125 consecutive cubes of the 
 natural series of nos. Using the plus sign, p=7, then .r-— 384, n~ 
 172, and by substitution we have 213 ... . 555, for the roots oi 343 
 cubes which fulfill the conditions. Tf ^//- 2, then using the insiius 
 sign, y>— 11, we have 1331 cubes: and using the plus sign, p=13, 
 and we shall have 2197 cubes. Now to find an even number of 
 rubes, we have only to add to formula [1] the term in the .original 
 scries next to [^+(^—1)3*, viz., ('•<•-{-;/ ) s , and then we shall have 
 {fyl—} K ; - (2/>- -3// 2 -; „),/• (,r-{-^) 3 == 2^r i -f-3>w ,2 +(2« 3 +«>+^ : ^ 
 U cube. But 2n must also be a cube. Let 2>/=y/; then n=--lj>'', 
 
 3;/ /> 9 -f2// 3 /v 9 
 and by substitution, we have /' :! ' :{ -j -n - r '~r 7 — .r-}-- ^-= a cube, 
 
 which, after dividing byy/', and multiplying by 8, becomes 
 
 ,sr 3 _;-i27 2 -f (2y/-f-4).r+^> 6 r= a cube, which iet==(2x+j0*)*. Redueing 
 
 p«— 3^+2 (^— lf-3 (y>-l)^ + l)2_ 3 
 
 llus we find .r= ., ., — tv= * — a 1 where, 
 
 6(y> 2 — 1) 6 6 
 
 in order to have x integral, p may be taken any even number, ex- 
 cept 6, and its multiples. Let p— 2 ; then ./— 1, u=4. Substituting 
 these values in the original series, we have —2, — 1, 0, 1, 2, 3, 4, 5, 
 four of which numbers balance one another, and one is 0; whence 
 we have 3, 4, 5, the roots of three cubes whose sum is 6 3 . 
 
 Let f> 4; then .'- 37, m 32, ;i iid \>y substitution we have 6, 7, 
 8, 9, . . . . 66, 67, 68, 69, for the roots of 64 cubes answering the 
 conditions of the problem. Let ^>— 8; then a— 661, n— 256, and 
 we have the series 406, 407, .... 917, for the roots of 512 consecu- 
 tive cubes of the natural series of numbers. Let/>— 10; then x — 
 
 1633, ?<=500, and by substitution we have 1134, 1135, 1136, 
 
 2132. 2133, for the roots of 1000 cubes that fulfill the conditions. 
 
10 
 
 6. To find /t square numbers such, that if each be either increa- 
 sed or diminished by its root multiplied ,by some number, the re- 
 spective sums and differences shall be squares. 
 
 Solution. We have here to make tfj&umm Q, if ±:by=\~}, 
 
 ^ = h : r;=[3i, &c. ad infinitum. Let .)^-{-rr.) t =[J i =m 2 .r 2 i then will 
 
 a ,,,.., <? " 2 r-i 
 
 a~^i— , ' and, by substitution, ■"-—'(•'— ^ /(2 _ l y — / ^:i =U- 
 
 Dividing this by <('\ and multiplying by (/>/-' — l)* 2 , we have 2 — < 
 □, which is so when warfci, Let i«a=*-l, then 2-i» a =^V+S« — >r 
 
 =Q— (l— — " ) • Keducinir, we find rc= ~ 7 '/' ,--.?-; whence, 
 
 w r ' /, «■ j and «s=— 5 — ,-- , / •> — 1\5 whence, a may bo 
 
 taken =Mp£{ 7-— // J ), and then arte (7/-; -£*)*. Let p=- 1 and //=--- 2 ; 
 then a $4, .<• 25; whence, #±«W£==0£5±6OO==85* or 5*=0«- 
 Again: let p=2, £~=3, then fc=l20, #=169, which let=£, and ,y, 
 
 respectively; whence, // 2 zb/>//— 28561 ±20280— 221 2 , or 91 2 =Q 
 
 Also, let/>— 1, 7=4, then </=240, a?=28&, which let=0, and a, re- 
 spectively; whence, *♦+:<*== 83521 ±69360— 391 2 , or 119 2 =[ ;'s, and 
 so on, ad infinitum. 
 
 7. To find a common value of a; that will make a^rbaaJwsQ, 
 
 x*zkibx= , y J ±r.r— j j, &c., ad infinitum. 
 
 Solution . In order to solve this problem, it will be necessary 
 to premise the following Algebraic Theorems from Barlow's The- 
 ory of X umbers : 
 
 1. Every. prime number of the form 4£+l, is the sum of two 
 squares in one way only. 
 
 2. The square of every such number is also the sum of two 
 ill one way only. 
 
 3. The product of any two such numbers is the sum of two 
 squares in two ways; the product of any three such numbers is the 
 sum of two squares in four ways ; the product of any four such num- 
 bers is the sum of two squares in eiu-ht ways; and, generally, the 
 product of any {n\ such numbers is the sum of two squares in 2 n ~ 1 
 ways. By attending to these theorems, the following solution will 
 be readily understood. 
 
 Put ,< = --; then by substitution, and expunging z 3 , we have 
 <? 2 db'->==Qi ^ 2 zfc^-=[j, **±<"=0> &C. ad infinitum. 
 Let f—ttf-^-ti 1 , and a=2mn\ theti« 2 db«^= s »* a rfc2wwi-|-w j ^=(m±n)*. 
 Also let ;r=p 2 -\-q\ and b=y>q; then z l ^zb=p 2 ^z2pq- ] r (/-=(p^zqY. 
 Again, let ??—i a -\-s\ and c=2rs; tlum ^ 2 ±c=;- 2 ±2r.s + .$ 2 -=(r±s) 2 ; 
 all Q's; and so on, ad infinitum. Here all the conditions of the; 
 problem will be satisfied, if we can divide r' into two squares in 
 any // number of ways. In order to do this, it will be necessary 
 to solve the following problem: u To divide a given square number 
 into two other square numbers''. See Baklow, p. 4(50. 
 Let A 2 =2 2 -=the given square, and /■'-', m\ be the rcquivrd squares'; 
 
 then A 2 =>; 2 4-"' 2 , and A 2 — w? 2 .=i? z . Assume A 4- >/*—-—-' and A- 
 
 1 . 9 <J 
 
II 
 
 n'r J>'r ,/r ( //' -\-</' 2 )r 
 
 ,c= K-. Hence we readily find 2A== S I 77--= ' ' ' aiul 
 
 <2,r=+— —,=— £y-j and thence we get P= Z£ l. ,/» '. «?=== 
 
 w ,., / ... — : in which formula. A, or its equal, z, may he taken anv 
 
 prime numher, or product of two or more prime numbers, each of 
 which is the sum of two squares ; then p'' 2 + </ 2 being pttt suecess- 
 ively equal to each sum of two squares into which A or z e:m be di- 
 vided, we shall have as many integral values of v, w ; whence also, 
 z—r-^-u-- the same number of different ways, for which may he put 
 successively, m f + ft*, ;/- + y 2 , r z -\-s 2 , &e. The different values of 
 2vw may he put successively=2>//?/, 2pq, 2ra, ©tc.,=€R, b, c, etc. 
 
 Let A=t*s=5==2 2 + 1 4 , and put ;/— 2, <j-=-\ ; then /'— 1', ?/?==3, and 
 ,r^=5 2 =4- + 3 2 , being the sum of two squares in one way only. 
 
 Let A=^=3.13=65; then 5=2 2 + l 2 , 13=3 2 +l"-\ 65=8-+ 1 = 
 7 2 + 4 2 , and putting^/, </, successively=the root of each set of two 
 squares, we shall have r— 52, to=$% W=60, w= 25, /• 1(5, «?— 63, 
 /• 56, ?«,•— 33; whence »*= -the sum of two squares in four ways. 
 
 Let ?=5.13.J7 H05; then 5=^2 2 +l 2 , ] 3 3M- 2 2 , 17 4-+1 2 , 
 5.13=65=8 2 +1 = T-+4 2 , 5.17=85= ^9 2 + 2 2 = 7 2 + 6 2 , 13.17= 221 = 
 lP + 10 2 =14 2 + 5 2 , li05=33 2 + 4 2 =32 2 + ^=-3l 2 i-12 2 -24 2 + 23 2 >; , 
 and putting y/, </, successively=the roots of each set of two Q's, 
 we shall have the following values of v, it, viz., 884, 663; 1020, 425 ; 
 520, 975; 272, 1071; 952, 561; 468, 1001; 1092, 169; 1100, 105; 
 700, 855; 264, 1073; 576, 943; 744, 817; 1104, 47: whence;-' the 
 Bum of two squares in thirteen ways. 
 
 In like manner, if «=t&e product of femf factors, each of which is 
 the sum of %wo squares ; then, taking the factors singly, tbe pro- 
 duct of every two, the product of ev*»ry three, and finally, the pro- 
 duct of all four, we shall &n&forty ways in which z* will be divided 
 into the sum of two squares ; and if 2=ithe product of live such fac- 
 tors, we shall, by proceeding as above, find one hundred and twen- 
 ty our. ways in which z 2 will be divided into the sum of two Hj's. 
 
 Premising that the subscript figures represent the No. of factors 
 used, we have the series 1^, 4 2 , 13.,, 40.,, 121 5 , 361,., 1093,, 3280 K , 
 occ, which may be extended ad infinitum, by multiplying any term 
 in the series by 3, and adding 1, to find the next succeeding term. 
 
 Having thus found the values of z, a, />, c, &c, in the expressions 
 z z ±a=[^], z 2 ±:b=^ 2 2 ±c=[], <fcc, if we multiply Attfte expressions 
 by X--Z 2 , we shall solve the original expressions 
 
 x 2 ±ax=[J, x 2 ±bx=^ry fe 2 i£eafc==0, <fcc. 
 
 Note. If 2=5.5.13= 325, then 5=-2 2 +l 2 , 13- 3 2 + 2 2 , 5.~> 25 
 4 2 +3 2 , 5.13 o5=8 2 +l 2 =:7 2 + 4 y , and 5.5.13=325- 1 8 2 -j V 17-| 
 (J- 1« 2 +10 8 , The last set is in the same ratio as 3'' + 2-, and 
 will give the same values of >\ w\ therefore, there are but 7 ways 
 in which z 2 can be divided into the sum of two squares. 
 
 We see from the above solution how to construct expressions of 
 this kind to anv extent we please, so as, having -riven the values 
 
12 
 of a, b, <\ etc., to liiul "-, and thence .<■. 
 
 8. It is required to find any n square numbers such, that if the 
 root of each he either added to or subtracted from the respective 
 squares, the sums and differences shall be squares. 
 
 Solution. Let ay±ax==0, W±fty=Ci ch*±ez=C\, etc., 
 ad infinitum. Here we will solve only the case a*sc*±aa5=Q; I'ut 
 
 </V'-L^/-— [n=mV J , then reducing we find x= 8 -$■, which be- 
 
 ingput in the expression a-y J —ax=[J, gives ^^ r-^-^,. 
 
 □• Dividing this by a ', and X'mghx (t/r — a' 2 )'\ we have 2<r — t ,f 
 2J ; which is so when m — -±:a. Let m=n — a, then, by substitution, 
 
 we have (c- r 2«)t-)t= Q= Ur — -^-n J 5 whence ?;*— — , , tf > 
 
 where a may be taken— /> 2 -|-</ 2 , and sr—2pq~ r 2q' J ; whence w /.---« — <y, 
 or — n—~p l — 2pq — q% and ,v= 
 
 P 2 +9* _ ff'+g' and , fJ ._ (/>'+?')' 
 
 ^?V — 4/yty \pq^f — /> 2 )' l ±pq{q~ — p 1 ) 
 
 Now let p=l, £=2, then «.=5 ; &=& <»==& (|f) 2 ±f!- (jj-J)' 
 and (fa)*. Again, let /?=2, #=3; then a=13, afe-j*^, which let - /> 
 and ^, respectively; then fty=«fc and (|M) 2 ±i|^(fM) 2 , fafc)* 
 Also, let p=l, q—t, then «.=17 ; as=-gf$r, which let==e and «, respec- 
 tively; then e*=fji, and (Ht/±|H=.(IH) , : aild («*ft ;m<1 so °°- 
 
 8. To find a common value of ./• that will make 8 aJp±rdK*==O, 
 £V\± &*:=□, ^r'4-c.y— [], etc., ad infinitum. 
 
 Solution . Dividing these expressions by u\ l>\ c\ etc., respec- 
 tively, and writing a\ b\ c', etc., for the reciprocals of a, b, <\ etc., 
 respectively, we iiave aJ 2 zha'aJ=:Q, qc?±-&x±=[J, «*rd=c'.r— [j, etc. 
 
 These expressions are of the same form as those in problem 7, the 
 only difference being that the values of a, 6, c, etc., are integral in 
 7 ; and a\ b\ r\ etc., in this, are fractional; x being integral in both. 
 
 Let -.r—z*; then substituting and expunging z\ we have 
 
 z iJ ±a'=[J, a£dr#==G, tfrfc^HH etc., ad infinitum. 
 Let .i- «,-r y/'-', :hhI ^''-^2^» ; then ^-±V=m 2 ±;2wy/- r vr=(///- ; //}\ 
 Also let z~p Q -~' r q'\ and b'—2pq', then z 2 ±.b , =p l ±.2pq-\~q' 1 —(p 
 And let sfe**-^ and c=2>-6'; then z 2 ±c'=r 2 ±2rs+s 2 =(r±s)\ etc. 
 
 Here, as in problem 7, all the conditions will be satisfied, if we 
 can divide ** into two squares any (n) number of ways ; a method of 
 doing whiph has been given in the solution of problem 7. 
 
 Let us subjoin a few examples of problem 7; as this is easily 
 changed to 9, they will illustrate both. See problems 11 and 12. 
 
 (a). Let 2=5.5.13=325; then, by Theorems on p. 10. we have 
 
 5=2*-j-l*, 18=3*+ 2 s , 5.5=25=4*+ 3\ 5.1 3=65=8*-- 1*=7*+4*, 
 
 5.5.L3=325 = 18*+1 2 =17 2 +6 2 =15*+10*. This last set being in the 
 
 same ratio as 3*+2 2 , as before stated, gives the same values to <\ «\ 
 
 Let/>'=2, <y'=l; then p=200, *r=195, and 2<.w>=101400=«. 
 
 • ' //=^3, q'=)>; ' ' /'= 300, ^=125, ' ' 2 /•//-- 75000=- b. 
 
 1 » jfe=^ 4=*i\ ' ' r=312, ?/r=91, ' ' 2cM=56784=- & 
 
 ' i //-. 8, ,/-. I; ' ' 0=80, ut=315, ' ' 2ow - 50400='/. 
 

 13 
 
 ' ' jt/=7, q'=4; ' '#=280,^=165, ' ' 2yw=92400=6, 
 > *» jt/=18, £ '=1 ; ' ' #=36, w=323, ' ' 2#w=23256=/; 
 1 ' p'=\1, q'=6; ' ' #=204, ^=253, ' ' 2vw=l 03224=*/. 
 . •. z 2 dz<:6=(325) 2 :±:101 400=207025 and 4225, or (455) 2 , (65)', 
 z-±b=(325Y ±7 5000=180625 » * 30625, ' ' (425) 2 , (175) 2 , 
 £ 2 ±;C=(325) 2 ±56784=162409 ' ' 48841, ' ' (403) 2 , (221)*, 
 z*±d=(:y25y ±50400=156025 ' ' 55225, ' ' (395)*, (235)*, 
 2 2 = te=(325)*±92400=198025 ' ' 13225, ' ' (445)*, (llo) ! , 
 2*±/=(325) 2 ±23256=128881 ' ' 82369, ' ' (350)*, (287)*, 
 z 2 ±<7=(325) 2 ;±103224=208849 ' ' 2401, ' ' (457)*, (49)*. 
 By considering this solution we readily see how to construct the 
 following problem: 
 
 (b). Make x J ±10n00x={J, ie 2 ±;75000.?;=n, x*±56784x=[J, ^± 
 50400*=n> «*±92400a;=O, # 2 ±:23256.£=n, «" ±1 03224x= Q 
 We see also how to find the value of a, when a, b, c, &c, are given. 
 
 Take any of the given quantities, as, e. g., 23256; then we have 
 2#w=23256, and #?c=l 1628=2.2.3.3.17.19. Put #=2. 2. 3.3=36, & 
 ^=17. 19=323; tfoen 2*==t>*+«0*^36*+323*=325*, which No. will 
 satisfy all the conditions; whence x=z~ is known. 
 (C). Let ;— 5.5.17=425; then 5=2*-j-l*,l 7=4*+ 1*, 5.5=25=4 2 +3 2 , 
 5.17=85=9*+2*=7*+6*, 5.5.1 7=425=1 9*+8*=16*+l3*=20*+5\ 
 The last set being in the same ratio as 4*+l 2 , will, as stated in the 
 Note on page 11, give the same values of #, u\ 
 
 Letjt?=2, ?=1; then #=340, #=255 and 2tfw?=l73400=</, 
 #=200, w=375 ' r 2mc=\50000=b, 
 #=408, #j=119 ' ' 2vw= 97l04=c, 
 O=180, tt>=385 ' ' 2utc=138600=tf, 
 #=420, ic=65 ? ' 2vw= 54600=6% 
 #=304, #-=297 ] } 2^=180576=/, 
 #=416, w=8! ' ' 2w=72384=y. 
 
 . ♦. z 2 ±;6r,=(425) 2 d= 173400=354025 and 7225, or (595) 2 , (85) 2 , 
 z 2 ±;£=(425) 2 :±150000=330625 ' » 30625, ' ' (575)*, (175) 2 , 
 ^ 2 = i = c=(425) 2 d=97l04 =277729 ' ' 83521, V (527) 2 , (289)*, 
 * 2 z h6/=(425)*±;138600=319225 ' ' 42025, ' ' (565)% (205) 2 , 
 S±e =(425) 2 zb54600 =235225 ' ' 126025, ' ' (485) 2 , (355) 2 , 
 2*±:/=(425)*±:180576=361201 ' ' 49, " (601)*, 7 2 , 
 2: 2 d=r/=(425) 2 ±72384 =253009 ' ' 108241, 7 ' (503) 2 , (329)'. 
 From this solution we easily construct the following problem : 
 (d). Make x 2 ±;173400.r=Q .r 2 ±150000.r=Q, .f 2 zb97l04.r=Q, x*± 
 138600r=Q, ^±54600=0, .z 2 :±180576.r=Q and « 2 zb72384^=n. 
 We see, also, how to find the value of x, when «, b, c, etc., are given. 
 Take any of the given quantities, as, e. g., 97104; then 2vw= 
 97104, #^=48552=2.2.2.3. 7. 17. 17. Put ^=2.2.2.3.17=408, w= 
 7.17=119; then * 2 =# 2 + #y=(408) 2 +(ll9) 2 =(425) 2 , which number 
 satisfies all the conditions simultaneously; whence x=z* is known. 
 
 Hence, also, Ave see how to solve the problem, 
 (el Make .<•'— (101 400) 2 =Q a 2 — (75000) 2 =n, x l ~ (56784) 2 =n. 
 
 Here, r 2 —(l01400) 2 =(.r+101400)(x— 101400), x 2 — (75000) 2 = 
 (;e+75000)(a;— 75000), and z 2 — (5 6784)* =(*+56 784) (x— 56784). 
 
 Let x=z*; then it is evident that we have to make 
 * 2 ±101400=Q z 2 ±;75000=Q * 2 :±56784=g ; for if each factor is 
 a 0> the product is also a QJ. The problem is now the same as (ft). 
 
 p'= 
 
 =4, 
 
 '/= 
 
 = 1 
 
 i ? 
 
 p'-- 
 
 -4, 
 
 <i'= 
 
 =3 
 
 j ? 
 
 p'= 
 
 =9, 
 
 </'= 
 
 --2 
 
 j i 
 
 p'= 
 
 *1i 
 
 q'= 
 
 =6 
 
 f ? 
 
 />'= 
 
 19 
 
 , </ 
 
 =8 
 
 > ? 
 
 r_ 
 
 — 1« 
 
 t 
 
 — i ' 
 
 ? ' » 
 
14 
 
 10. Find three square numbers in arithmetical progression, 
 such that if from each its ro6t be subtracted, the three remainders 
 shall be rational squares. 
 
 HISTORY. So far as we know, this problem was originally published In the first London edition 
 ot J. R- Young's Algebra: where erroneous answers were given. The American Editor, noticing 
 this fact, omitted the problem ; and In a note he says, "It would be difficult to find true numbers of 
 an} mock-ration. uleM negative answer* be admitted". 
 
 Solution . Let "'•'< />'•''% '''-''% represent the numbers required. 
 Then, by the conditions, [tf^— axaJfc} •• • ( 1; )i S** 2 —- &*=0 • • • (2), 
 and chfc--<&h==C] . . . (3)] . . . (C). Put a?a?— aoF=^n&& ; then will 
 
 x ~— — — i - . . . (4). Putting this value of x in (2) and (3), and 
 
 multiplying by (// 2 — m~)'\ they become, respectively, 
 
 a-l>' x —X<r—»r)u!^ Q . . . (5). and oV— (a a — m 2 )ac^Q . . . (6). 
 These are QJs iftf?— -fca, but either of these values of m renders x 
 infinite. If n±:a be substituted for m in (5) and (6), we have 
 a t b t ±.2aH>a-\- abn' l =-'~} . • • (7), and a*c 2 ±2a 2 cn+am*=={3 . . . (8). 
 
 Take ah~np for the root of (7), and we have n ,— •>_s =r i ) — • •♦ ( 9 )* 
 
 (rr — ab \ t 
 n_ — \ 
 
 we obtain <*p*±£al><p i +2a6c(2a+2b-- c)j^±4o*^+V*V==C] . . . 
 (10). Assuming cjZ+iabp—ttfc for the root of (10), we have 
 
 2abr ac-\-bt — ab , . 
 
 ,i or "^ • • • (11)- 
 
 1 a(> — ac—Oe 9 2e 
 
 But if (p* — tabp — abc be taken for the root of (10), we have 
 
 ab — ac — bo 2abc . 
 
 pz=^ or . , 7 ... (11 ). 
 
 1 2 c ac-\-bc — ab 
 
 From (4), *F^ a t_ n f i or, since m*=a t ±2an+n*, a?=— J^ggj 
 
 . . . (12). If we substitute for n in (12), its value as show r n in (9), 
 we shall get 
 
 4aop{pzta){p±:b) ' ' ' l ; ' 
 If, in (13), we substitute for p its first value in (11), or that in (11'). 
 using the -j- sign in the binomial factors of the denominator, we 
 
 shall find 
 
 —{ {a.b—ac—bc)*—iab(? j> 2 ( 
 
 x ~ ~Sabc(ar—ab—bc)(ab—ac-i-bc)(ab-\-ac—bc) ' ' ' ' * 
 
 When the relation between a, b, c, is such as to make the numeri- 
 cal value of a? in (14) negative, (C) becomes [a l x 2 -{-ax=[^ . . . (l')i 
 b*x*-\-bx=[J . . . (2'), c»x l +cx= D • • • (3')]-- • (C); then 
 
 <! (ab—ac—bey—labc* y . 
 
 x Sabc(ab — ac — bc)(ab — ac- j r bc)(ab-{-ac — be) ' ' ' } '* 
 This shows that for such positive values of a, b, c, the conditions 
 of (C) are impossible, by the general method, when positive an- 
 wers are required, and that (C) must be changed to (C), that the 
 problem may be possible; a-x 2 , fe 2 , c 2 x-, being three square num- 
 bers, such, that if each be added to its respective root, the sums 
 will be rational squares. 
 
 Whence, premising that A represents the square root of the 
 
15 
 
 numerator of (14), and B the product of the trinomial factor* of 
 the denominator, the required squares, in both cases, will be 
 
 a 2 x 2 ±ax= ("g^g) \a*&— 2a 2 be— 2ab 2 e— a 2 e 2 +2abi?— /r^)> 
 b 2 x 2 ±bx= (g^]gJ \a 2 b 2 — 2a 2 be— 2ab 2 e— SoV-f- 2a$& +-W% 
 
 _^-J (Sa 2 b 2 — 2a 2 bc— 2ab 2 c— a*t?+2a&c— 6V)». 
 
 Again: if, in expression (13) we put for jd either its second 
 value in (11), or that in (11'), employing the — sign in the factors 
 of the denominator, we obtain the same value for as; a, b, e, being 
 an)- numbers whatever. Since a, b, c, may be interchanged with- 
 out altering the value of K, it is evident that this process will 
 give but one value of x that, will fulfill the conditions in (C) ; but 
 other values of /> may be found by substituting 
 
 , ac A -be — ab 2abe „ . , . i . . .. 
 
 jftf- — 2~ > or Sr+^qp^^Z^g' for ^ in CIO), and making tho 
 
 result a square; the values of p t thus found, will produce other 
 values of x that will fulfill the conditions in (C). 
 
 Since the numerator of a? is a square, and the denominator divis- 
 ible by a,' b, e, the square of the numerator of x will be the com- 
 mon numerator of ax, bx, ex; therefore, when a 2 x 2 — ax, b 2 x 2 — bx, 
 c 2 x 2 — ex, are squares, if the denominator of ax, bx, ex, be subtracted 
 from the numerator, the remainder will be a square; but if a 2 ic 2 -j- 
 ax, b 2 x 2 -\-bx, c 2 x 2 -\-cx, are squares, the sum of the numerator and 
 denominator of ax, bx, ex, will each be a square. 
 
 To fulfill tjie conditions of the problem, a 2 x 2 , b 2 x 2 , <£a?, :>r, ex- 
 punging sb*, a 2 , b 2 , c 2 , must be in arithmetical progression, and have 
 such relative values as to make x positive. 
 
 a 2 =(p 2 — 2p— l)y, b 2 =z(p 2 -{-l) 2 (f, c 2 =(p 2 —2p+l) 2 (j 2 , are in 
 arithmetical progression; and to obtain the least numbers for the 
 numerator and denominator in the fractional value of x, let p=*, 
 and q=l; then a=31, b=25, and c— 17. Or we may take 
 a 2 =(r 2 — 2rs— s 2 ) 2 , b 2 ^(r 2 -\-s 2 ) 2 , c 2 =(r 2 -f 2rs— s 2 ) 2 ; in which the val- 
 ues of r-\~s, r — s, being substituted for t\ 8, give the same values 
 of a, b, c, or the same multiple of those values; and any multiple of 
 the values of a, b, c, will give the same value of & If r= -4, s— 3, 
 then will a= — 17, 6=25, c=31; but if r=1, and 8=1, then will 
 a— 34, #=50, c=62. Substituting either of these three sets of 
 
 , , . • /1/A wj (864571 )* . 
 
 values for a, b, c, in (14), we find *^ n 01 1044 931800 * 
 
 . . (864571) 4 • ., , /864571.315871V 
 
 * ' * a2aj2= (647708525400?' and ^~^== { wi 08523 too) '> 
 
 2 _ (864571) 4 / 864571.554113 X \ 
 
 b f ~(440441797272) 2 ' bx ~0x— ^ 440441797272/ ' 
 
 2~2- 
 
 C'X 
 
 (864571)* 2 2 / 864571.026329 N 
 
 (3551949978U0)-' c x —ex— ^ 355194997800/ 
 
 These are the least numbers that have yet been found, when the 
 roots, ax, bx, ex, are taken with the positive sign; when reduced 
 
16 
 
 to a common denominator they become 
 
 2 2 _ / 12707211238697 \ * ' . 2 2 __ _ / 4642579407797 \ 2 
 
 a '— Vj 1011044931800/ ' and a x ^—^11011044931800/ 5 
 2 _ / 1868T075351025 \ 2 2 _ _ / 11976750763075 \ 2 . 
 
 ° X ~ A11011044931800/ ' ~ V11011044931800/ ' 
 
 2 2 _ / 23171973435271 \ 2 2 2 __ , _ / 16786682585629 \ \ 
 
 c2 ^—\l 1011044931800/ ' C X ~~ ex — ^11011044938100/ 
 
 The product of the sum and difference of two quantities being 
 equivalent to the difference of their squares, the numerator of (14) 
 may be exhibited under a different form, which has the advantage 
 of symmetry, ot corresponding* more nearly with the denominator, 
 and of showing, at a glance, that «, b, c, may be interchanged, but 
 the change of form introduces radicals; all of which, however, dis- 
 appear upon performing the multiplications indicated. We have 
 (ab — etc — be) 2 — ±abtr=(eib — ae — bc-\-2c\/ab)(ab — a e-bc-2c}/ ab)= 
 
 \ ab—(ac—2cyei,b+bc) }-.<{ ab^-(ac+2c-[/ ab-\&c) y = 
 ■{ ab—i^/ac—ybe) 2 )■.-{ ab—(^/ac-\ \/be) 2 )■ =(y 'ab—y 'ae-\-y be) 
 {yeib+}/cic—y / bc){y / ab—y / eic—y / 'bc)(y / 'cib-\-y / ac J r }/bc). 
 
 11. If, instead of three □ Nos. in arith. prog., we should wish 
 To find three □ numbers, <&x\ b*x\ cV~, such, that (a 2 c 2 dza.r=[J, 
 5%**±f&t=0 cV-~ c/— D) • • • W, the value of r, in [14], fulfills 
 
 three of these conditions; and [13] gives ac= __ 4 ^ a _^ (f>+ p ) » 
 
 the numerator of which being a square, we have only to make the 
 sum of the nuiner. and denom. a □, in order to have a?x 2 — «#=["]. 
 Since the denominator is negative, the sum of these is found to be 
 p*— 4tbjt— Gabp*— W*p 2 — ±ab-p+aVr= [J . . . [16]. 
 
 2a b 
 
 Taking^ 2 — 2bp— ab for the root of [16], j?=— g , , ^ • ■ . [17]. 
 
 ab — ae—bc 2ab ab — ae — be 
 But from [1 1 ], />=*= ^ ? . • . -£^== jj^ - x 
 
 and c= '., T , /., • • . [isX Putting this value of e for c in [14], 
 a- — ab-\-2b- L J ° 
 
 multiplying by b, and dividing both numerator and denominator of 
 
 the right-hand side of the resulting equation by 16« 6 & 6 , we obtain 
 
 (ar-\-4b 2 ) 2 
 bx=, — , „,v , w ' — , .,„. o - Since the numerator of this frac- 
 (a-\-2b) X — 4( — 2a-\-4b z )a 2 
 
 tion is a square, we have only to make the difference between the 
 numerator and denominator a Q m order to make b 2 x 2 -\-b.r=[J. 
 This difference is — 7« 4 -f 40a 2 b 2J rWb\ which must be a Q 
 
 If 5=1, this becomes — 7a i +40a 2 +16=[J • • • D 9 ]- This is a D 
 when a=±2', let, . ' . a=/-f2, then [19] will become-7/*— 56Z* 3 — 
 128/ 2 -64/-f 64=n, which put=(9/ 2 -4/+8) 2 ; then /=-}$, and <i= 
 /_|_2= T 6 T ; but 6=1, .'. a : b :: T \ : 1 :: 6 Ml. We may, there- 
 fore, take «=6, and 6=11; then, from [18], <?=-%£*. Multiplying 
 a, b, c, by53, to obtain integral values, a— 6.53, ^=1 1.53, c=7.66. 
 
 85 80 2 715 2 
 
 These values of a, 6, c, in [14], give ^= &O 4 78 4 5O72 == 1*2214688- 
 
 But if a=f-2, [19] becomes — 7/*+56/' 3 —l28/^+64/+<34==n, 
 
17 
 
 which put— (9/'*— 4/— 8) 2 , theii/=4f, a=/— 2= — &. Therefore, 
 a : h :: — ^L : I :: — 6 : 11. If a=— 6, 5=11, and from [18] 
 c= — H'i^' Multiplying a, 5, c, by 43, we may take a= — 6.43, 5= 
 11.43, c=— 11.12, and from [17],^=—- 3ii£A* These values of a, 
 
 65 2 
 (6, y>, in [13], g^yc x=-—~—-' Either of these two values of i 
 
 fulfills the five conditions in [6], 
 
 Other values of x may he found by writing v±j\- for a, in [19], 
 and making the results squares. Similarly, others may be found. 
 
 12. If it were required to find a common value of x, so that 
 //V 2 dz </.'■=□, b 2 x 2 ±zhx=[J l , c*x*±cx=[2, &c, ad infinitum, Ave re- 
 mark first, that all the prime numbers contained in the formula 
 U-\-l, are each the sum of two squares; which call r 2 -j-6' 2 / 7'' 2 -\-s' 2 ; 
 r" 2J \-s"'\ &c, successively, where r, s; r', s f ; r", a", &c, may be 
 any Nos. the sum of whose Q's is a prime, derived from \t-\-\. 
 Then take a/=[(r 2 +s 2 )(r /2 +^ 2 )(/ ,2 -f-s ,/2 ).&c.].(r 2 -|-s 2 ) 2 , 
 
 ^'=[(^4-^(^ 2 +^ 2 )(^ /2 +^ 2 ).&c.].(r 2 -f^) 4 , 
 ^"=[(r 2 +6- 2 )(/ 2 H-s ,2 )(/ ,2 +^3).&c.].(r 2 -fs 2 )«, &c, to 
 [(r 2 +^ 2 )(/ 2 +^ 2 )(r" 2 -f5" 2 ).&c.].(r 2 -f,9 2 ) a ». 
 
 Taking the factors r 2 -j-s 2 , r' 2 -\-s' 2 , and putting r=2, $-=] ; /=3> 
 .s'_rrr2, these general expressions Avill be ^=(2 2 -(-l 2 ) 2 (3 2 -|-2 2 )' 2 , x== 
 65 2 (2 2 +l 2 ) 2 , a ; "=65 2 (2 2 +l 2 ) 4 , aj" ; =65*(2 2 +l 2 ) 6 , &c, to 65 2 (2 2 -|-l 2 f». 
 
 As the no. of ways in which r 2 -\-s 2 can be divided into the sum of 
 two □'» is infinite, Ave can find as many expressions like the first 
 two in [6], p. 16, as we choose, all having a common value of x. 
 ^ Since 65 2 =(8 2 +l 2 ) 2 =(7 2 4-4 2 ) 2 =5 2 (3 2 -f2 2 ) 2 =13 2 (2 2 4-p) 2 , we have, 
 for four sets of three square numbers in arithmetical progression, 
 (8 2 -2.8.1-l 2 ) 2 , (8 2 +l 2 ) 2 , (8 2 +2.8.1-l 2 ) 2 , com. diff.=4.8.l(8 2 -l 2 )— a, 
 (Y-2.7.4-4 2 ) 2 , (7 2 -f-4 2 ) 2 , (7 2 +2.7.4~4 2 ) 2 , ' ' =4.7.4(7 a -4*)=0, 
 
 5 2 (3 2 -2.3.2-2 2 ) 2 , 5 2 (3 2 -f 2 2 ) 3 , 5 2 (3 2 -}-2.3.2-2 2 ) 2 , ' ' =5 s .4.3.2(3 a -2 2 )=c' 
 13 2 (2 2 -2.2.1-1 2 ) 2 , 13 2 (2 2 +1 2 ) 2 , 13 2 (2 2 +2,2.1-1 2 ) 2 , ' = 1S*A.2A(2*-V)=<?. 
 
 Writing for a, fa c, d, the square of their reciprocals, Ave obtain 
 
 4.; -- / 65 2 V 65 2 /65.89V /65.23\ 2 
 
 /•> . . a r-i f 65 2 \ 2 65 2 /65.79V /65.47N 2 
 
 W^«^P^ X 3696 \ ±3696= V 3696 J ' OT V 3696V ' 
 
 , f . _. / 65 2 Y 2 65 2 /65.85V /65.35N 2 
 
 w _ t _ _ /65 2 V 65 2 /65.91V /65.13V 
 *^mQ?T (4056) ±4056= ( 4056.) ' * ( io5e) ' 
 
 The above values of a, #, c, substituted in formula [14], page 14, 
 will not give #=65 2 ; the value of x, thus obtained, will satisfy 
 only three of the conditions in [6], 
 
 Because m 2 +w 2 +2m?i=Q, and m 2 -\-n 2 — 2wm=Q avc know that 
 w , +w , ±2mw=[], Avhich seems to be the base of the general prob- 
 lem to which these examples belong. Substituting r 2 — s 2 for m, 
 2rs for n, we have (r 2 +.s 2 ) 2 ±4r*(r 2 ---s 2 )=Q; an expression repre- 
 senting the extremes of three [] numbers in arithmetical progress- 
 ion, (r 2 -fs 2 ) 2 being the mean. Multiplying the last expression by 
 2 -K 2 ) 2 , it becomes (r 2 +s 2 ) 2 (r 2 +s 2 ) 2 ±4rs(r 2 — s 2 )(r 2 -{-s 2 y={J. 
 
 Writing x, for (r 2 +s 2 ) 2 , Ave obtain x 2 ±_±rs(i' 2 — s 2 )x=[J . . . [c]. 
 Dividing [#:] by the square of the coefficient of x, and putting #, 
 for the reciprocal of its coefficient, we get a 2 x 2 ±ax={^\, and so on. 
 
18 
 
 13. Make «W-\-dx=-U • • • [1], &tf$r&*=Ci - - . [2], and 
 cW-rfc=a . . . [3] . . . [A]. 
 
 The solution of the above case of triple equality produces a 
 formula so extensive in its application to the solution of those 
 Diophantine Problems which require the fulfillment of six or more 
 conditions, that we are surprised to find that no writer on Dio- 
 phantine Algebra has given it a place in his work. To the student 
 just entering upon the study of double and triple equalities, a knowl- 
 edge of theextent to which this formula can be used will be found 
 invaluable; but he will search in vain in the works of Eulek, 
 Baeloav, Lagrange, or any of those authors who have written on 
 this subject, for the information he requires. 
 
 Solution. Let «V|fcm¥; then will x= m% __ a % ' ' ' [4]. 
 
 Putting this value of x in [2] and [3], and multiplying by (nt 2 — a 2 )'', 
 they become, respectively, 
 
 b\P+(m 2 —a 2 )de==[J . . . [5], and c V?+(w a — a*)df=\J . . . [6]. 
 These are 0' s if m==b«, but cither of these values of m renders 
 x infinite. If n-\-a be substituted for m in [5] and [6], we have 
 hhl'^-^adeii-lden 2 ^^ . . . [7], and <*&+ 2adfn+dfn i = □ . . . [8]. 
 
 1d(biJ-\-ae\ 
 Taking bd-np for the root of [7], we have n= J_4 e ' ' M- 
 
 Placing tli is value of n in [8], & multiplying by {p 2 —de) 2 , we get 
 c*p 4 \<ihjy \ (4a 2 ef+ib*$f— 2c a ^)^*+4a5<^^+c z ^fe|J , . i [10]. 
 
 Assuming ep 2 — z<L ^-p — ede for the root of [10], and reducing, 
 
 a 2 lrf—a^e—Wd _ - _ rj _ d 
 
 P^ ^ /%a b$ -'"l u l From M. »=TSP5?S or > 
 
 since m^a*+2an-tr?i «= n(n + 2a) ' " : [#)• 
 Substituting for n in [12], its value, shown in (9), we find 
 
 x= t , ^T?j? , x * ' ' [13]. If in [13] Ave 
 
 \X>{ap-Ybd){bp-\-ae) L J L J 
 
 substitute for^?, its value, taken from [11), we shall obtain x= 
 
 <{ (a>b 2 f—a 2 c 2 e—b 2 c 2 cI) 2 —4a 2 b 2 cUk j> 2 % t 
 
 Ka 2 b 2 c ? (a 2 b 2 f-d\?e-bW ' " L J ' 
 
 a, b, c, d, e % /', having any values; but if their relative, values be 
 
 such as to make x negative, it solves the triple equality 
 
 «*x 1 —dx={3, W— *B*=Q c 2 x 2 -~fx=\J. 
 
 Therefore the foregoing work is a solution of the formula? 
 
 a 2 .<*±dx=[J, bW±ex=\3, (*x 2 ±.fx=U> 
 
 the double sign being taken disjunctively. 
 
 Writing a for d, b for 6, e for /, in \_A], it becomes 
 
 a 2 x 2 -\-ax=C\ i £V+&b=Q c 2 x 2 -\-cx^=C and [14] is transformed to 
 
 j (ab—ac—be) 2 —±abc 2 j> 2 . , . ^ e 
 
 8abc(ab—ac—-bcXab—ac-\-bc)(ab-^ac—bc) L J ' 
 
 same as formula [15], p. 14. 
 
 If x be negative, the sign of the second term in each of the three 
 
 formula? above will be changed ; therefore, when 
 
 — \ (ab — ac — be) 2 — kabc 2 } 2 f .. , , 
 
 . x ^8abe(ab—ac—br)(ab--aclbc){ab-tac~-b4 ' ' ' ] - bJ ' (tne 
 
19 
 
 same formula that [14], p. 14., ought to have been}, it fulfills the 
 conditions d l x l — aa;=|_J, b'W—bx^U, c l x 2 — ex— |_J. 
 
 [14] solves all cases <rf 'triple equality when the terms contain- 
 ing .r have the positive sign, and each of the three formula 1 consists 
 of" two terms only; one of Which is some square multiple of x'\ 
 and the other any multiple of s». 
 
 We subjoin a f ew of tlie cases, in all of which the double sign 
 must be taken disjunctively, because by this formula only one of 
 them can have place with the same coefficients of x" and x ; but 
 which of them is to be used depends upon the relative values of 
 these coefficients. 
 L tf^'aa*=tj, b\c 2 ±.bx=U, <Wdt(s»=CJ 
 
 2. a 2 x 2 ±bx={J, b 2 x 2 ±cx=^[3, c 2 x 2 ±ax=[J. 
 
 3. a 2 x 2 ±cx=Cl b 2 x 2 ±«x=\J, c 2 x 2 ±:bx=\~\. 
 
 4. aW±ixx==£% b i x 1 ±ax=\J, c 2 x 2 ±ax={J. 
 
 5. «W±(b+c)x=a Pgf±{a+ey^2> c*x 2 ±(a+b)x=:[J. 
 
 6. 'a&±(ajrb+c)x=13, fc*±(a+b+c)x==£\, *&db(p#b'+c)x=Q; 
 
 I. a 2 x 2 -±(b— c)x=\J, b 2 x 2 ±{a—c)x=\J, c*3?-t(a— M*=Q. 
 
 9. (a'+P^zttt^D, (^+C^«^±fe==Q, (^H-c 2 ).?- 2 ±;^.r^Q 
 
 10. (a»+y-fc»)a*±^=D, (^+^+'^±te=Cj; 
 
 II. {a 2 — P)x' 2 -tcx=[J, (a t —c 2 yx 2 ±bx=\J, (//— c 2 ).^zh<^— Q 
 
 12. #£{fl+i)«=P, ^#^C«4-C>=n, ^r 2 r 2 H-(/>+c)«r--- | 
 
 13. a 2 x 2 zbniax=E], tW±Mbz=£3 } c 2 x 2 ±pcx=\^], 
 
 14. « 2 x 2 ±m(a-{-b-\-c)x=[2, b 2 x 2 ±n(a~\-b-i-c)x=[J, 
 
 <k?±: p(a+b+c)x~iy 
 
 15. tt*5Vi(a— &)#=0 aWx*±(a— c)«=n, aWfc»=fc(&-- c)a 
 
 16. ^V> 2 c 2 .r 2 d=«£C=n, a 2 6V/ 2 ztfrr=:Q «*#W±e&=Q 
 
 17. «W^±(a-t-ft)a;==n, a'^i*i(^4-c)afe=d, 
 
 ^VV 2 ±(/>+c>«==--Q 
 
 18. (a-\-byx 2 ±ax=[J, (a-\-e) 2 x 2 ±bx=\J, (b-\-cfx 2 ±cx=r^ 
 1 0. (d -|-&)V:±(a+*)a;=n, OH ^) 2 ^=b(^+ c)a?=Qi 
 
 (&+<?)V±(H-<0*=Q 
 20. (a— ft) 2 ^±(a+A)a=D, («— c) a ^±(<B+c>==n> (&— c) s ^±(A 
 -J-c)jc=[] ; a, #, c, <?, e - , / ??<, ??, p, having any values, except in 0, 
 10, 11, where they must be such as to make the coefficients of x 2 Qty 
 
 Formula [15] solves No. 1, of the above cases. To solve No. 2, 
 substitute b for d, c for e, and a for/', in [14] ; then will 
 
 ■{ Q 3 £ 2 — fl 2 c 3 — //c 2 ) 2 — 4r<W V) 
 
 ***'*#&&{&&— €p^$j)& ' *[' ' I 
 
 To solve No. 3, write c for d, a for e, b for/, in [14] ; then 
 <{ (a>&—cP<?—h'<*f—\cPb*& y 2 J 
 
 To solve No. 4, put a for flj e; and/ in [14], and 
 
 ^TSIjW^I^^^ • • • [ i f>j. 
 
 To solve No. 5, write # |-c for tf, <7.-|-c for e, and <7-j-# for/ in 
 [14] ; then we shall have 
 
 { [a 2 bXa+b)—a 2 cXa- \-c )—b 2 c 2 (b-\-c)~] ?— 
 
 x ~Sa 2 b 2 c\d z b\a + b) —a 2 c 2 (a-\~ c) ~b' 2 c\b + e)].[a'0*(a-| 5) : ■<> ft -" ( ■ i 
 
 4«W(«+c)(S+c) V 2 ' _ r9nl 
 
 +W^(H-c)].[aW(a Vfy) I Sc*(S ! SJ &V(£+c)] L ^ J * 
 
20 
 
 To solve No. (5, writing a-\-b-\-c for d, e, and/, in [14], 
 .iiid dividing numerator and denominator by (a-\-b-\-cy, Ave get 
 
 \ {a 2 b 2 —a 2 c 2 —b 2 c 2 ) 2 ~\a 2 b 2 & y.(a+b+c) 
 x ~~Sa t b l c i {a 1 b 2 ~ a 2 &— b 2 c*)(a 2 b 2 — €?€*-{- b 2 c 2 )(a 2 b 2 -{-a 2 c 2 —b 2 c z ) " *L 21 > 
 It is obvious that if a, b, c, have such values as will fulfill the 
 several conditions of a Diophantino Problem, each of Avhich is ex- 
 pressed by homogeneous terms of the same degree in each, the re- 
 quired power corresponding to the degree of homogeneity; ax, bx, 
 and ex will also fulfill the same conditions, whatever value x may 
 have. Now formula? [15] .... [21], inclusive, show the values 
 of x that will satisfy the conditions of either of the equalities 
 above, numbered 1, 2, 3, etc., 0, b, c, having any values; therefore 
 the values of a, b, c, that will fulfill the requirements of any homoge- 
 neous formulae, substituted in [14] [21], inclusive, will 
 
 give such a value ofx that ax, bx, ex, will fulfill not only the con- 
 ditions of the problem whose terms are expressed by the homogene- 
 ous formulae, but also those of any one of the triple equalities above, 
 marked 1, 2, 3, etc. 
 
 To find three square numbers in arithmetical progression, sucli 
 that if each be added to its root, the sum shall be a square. 
 Substitute for h } b, e, in [15], the following formulae, respectively: 
 jr — 2pq — q 2 , p*-t-q' 2 , p 2 -\-2pq— q 2 , . . . *{-&)-• 
 The resulting value of x will be such that 
 
 (p*—2pq—q 2 y 2 x\ (jP+qytf, (pj 2 +2pq— q 2 fx 2 , 
 will be the required squares,^ and (/being assumed at pleasure. 
 If ^j>=4, ^—3, x will be positive; p=2, </=l, x will be negative. 
 
 14. If the three squares are to be in geometrical progression, 
 place ra for b, and r 2 a for c, in [15] . . . •{ (J \. 
 
 15. If they are to be in harmonical progression, write, in [15], 
 a=p* — 2p z q — 2p<f — p*, b=jj* — 6p\/-\-q\ c=p i -\-2p*q-\-2pq*—p i ; 
 
 being the reciprocals of the formulae in -{ B )■ . 
 
 16. To find three square numbers whose roots shall be in arith- 
 metical progression, and if to each be added the sum of the roots, 
 the sum shall be a square. 
 
 Substitute p~\-q for a, p for b, p — q for c, in [21]; or these values 
 for a, b, c, and 3pj for cl, e, and/, in [14]., 
 
 17. If the roots are required to be in geometrical progression, 
 write the values of a, b, c, as shown in <[ C )>, in [21]; or these 
 values for a, b, c, and their sum for d, e, and/, in [14]. 
 
 18. If the roots are to be in harmonical progression, substitute 
 />(p-\-q) for a, p 2 — q 2 for b, p{p — q) for c, in [21] ; or these values 
 
 for a, b, c, and their sum for d, e, and/, in [14]. 
 
 Let us show some of the applications to double equalities. 
 
 19. To find a number, such that if it be either increased or 
 diminished by a given number, a, and the result be multiplied by 
 the number sought, the product shall, in either case, be a square. 
 
 Let x= the number; then must x 2 ^\-ax=^\Z\, and x 2 ^ax=^=-\^}. 
 These correspond to [1], [2], in [yl], . • . a=l, b=l, d=a>, e=— 1. 
 
 Substituting these values in [13], we find as— . / 2 ^ ' 
 
 (iv afr 2 —\— < ii 2 
 
 Writing s for p in this, it becomes «••== 4y .^ r »_^»j ' ' ' t&} 
 
21 
 
 Since this value ofx renders x" — ax and x*-\-ax squares, we have 
 for three squares in arithmetical progression, x 2 — ax, se 2 , x 2 -\-ax. 
 Writing for x 9 its value, these become, after obvious reductions, 
 
 aXr*—2rs—sy, « 2 (r 2 +s 2 ) 2 , a 2 (r 2 +2rs— s 2 ) 2 . 
 If a=l, r=2, s=l, we have 1, 25, 49, for three such squares. 
 
 20. To find two square numbers whose sum shall be a square, 
 such that if to the first any multiple of its root be added, and from 
 the second the same multiple of its root be subtracted, the sum and 
 difference shall each be a square. 
 
 Let a 2 x 2 and b 2 x 2 be the required numbers ; then must 
 a 2 x 2 -\-max=\^], and b 2 x 2 — ??ibx=[J. 
 a 2 x 2 -\-b 2 x 2 will be a square if a=r 2 — **, b=2rs. Writing, in [13], 
 r 2 — s 2 for a, 2rs for b, m(r 2 — a*) for d, — 2mrs for ?, we find x= 
 
 <( p 2 — 2?n 2 rs(r 2 — s 2 ) )* 2 . 
 
 4p ■{ 2rs(r 2 — s 2 ) p 2 -\- [4mf'¥(r 2 — s 2 ) — 2?nrs{r 2 — s 2 ) 2 ] p— 4wi. W(-r 2 — rf \ 
 If the difference of the squares is to be a square, take a=r--\-s\ 
 b=2rs. If one of the roots is to be added to its square and the other 
 to be taken from its square to produce the required squares, m=\. 
 
 21. To find two square numbers such that each, when added to 
 their product, shall produce a square, if to the first its root be add- 
 ed, the sum shall be a square, and if from the second its root be 
 subtracted, the remainder shall be a square. 
 
 a=lSn, b—32n, satisfy the first two conditions; substituting 
 ISn for d, — 32« for e, in [13], the problem is solved. 
 
 22. To find three square numbers, sueh that the sum of any 
 two of them shall be a square. 
 
 Let m 2 (n 2 — l) 2 , (2mn) ! , w''(m 2 — l) 2 , be the required squares; then 
 the sum of the 1st and 2d is m 2 (?r+l)— D? 8um of 2d and 3d is 
 w 2 (m 2 -f-l) 2 =Q, of 1st and 3d is m 8 n s (m 9 +»*) — 4mV-fm s -rW J =[]., 
 
 Let ?n=pn; then, substituting and dividing by vr, we have 
 2^(]f J rl)n 4 ' — 47> 2 >i 2 -{-£>*" -fl==D> this is so when n=±:\. Let n= 
 q — 1 ; then substituting and arranging the terms, we have 
 P'(p*+l)(f— 4/> 2 (7> 2 H-l)V+2y(^ , + l)r— 4;Ay-%+0> -!)*=□• 
 Putting this=[(/?»— 1)— ^q+- .^ 1 V ]'» we find 
 
 4(;r— 1) , p 2 — 3 p(?r—l) 
 
 <l=^h> — 7^ '■> whence n=J— — -, m=±AL — _L . § tne roots ar0 
 Sp 2 — 1 Sp 2 — 1 op' — 1 ' 
 
 after expunging the common factor, or divisor, (3/r — I) 3 , 
 
 8^(^—1), 2p(p*— 3)(3p 9 — 1), (p>— l)(y+4p+l)(^— 4/? + l), 
 where p may be any number ^> 1. Let jp==2 ; then the roots of the 
 required numbers are 240, 44, and 117. 
 
 Writing a=44, 5=117, c=240, and substituting in [15], we find 
 a value of a;, such that a*x\ b 2 x 2 , e 2 x 2 y fulfill the conditions of this 
 problem and those in Case 1. The above values of a, b, c, substi- 
 tuted in [17], [18], .... [21], produce other sets of numbers. They 
 also solve Case 9, for they make squares of the coefficients of x 2 . 
 
 If Ave write «', b', c', for the coefficients of x 2 ; d, e, f, for those 
 of x, in [14], we obtain sets of numbers that satisfy any one of the 
 20 given Cases (except 10, 11), and we know not how many more, 
 because by giving different values to d 5 e, /', the number of Cases 
 might be increased sine limite. 
 
22 
 
 g^^^ 23. It is required to find four square numbers such 
 40HIm& that (1) the sum of their roots shall be a square, (2) the 
 R^sum of the first and fourth, (3) the sum of the second and 
 ^1^^^ fourth, (4) the sum of the roots of the third and fourth Xd 
 by the root of the third, or (5) by the root of the sum of the second 
 and fourth, (6) the sum of the first three added to three times the 
 third, (7) and the sum of all four added to three times the third, — 
 shall all be squares ; (8) the product of the third and fourth added 
 to the product of the first and second, (9) to the product of the first 
 and third, (10) to the product of the second and third, or (11) to 
 the sum of these three products, — shall produce squares ; (12) any 
 one of the Nos. added to the sum of the roots of the other three, 
 (13) the square of the sum of any two of their roots added to the 
 sum of the other two roots, or (14) the square of the sum of any three 
 of their roots added to the remaining root, — a square; (15) also, 
 four other square numbers such that either of the first three added 
 to the sum of the roots of the other three shall all be squares, and 
 which also shall fulfill conditions 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 
 
 Solution. Let a 2 x 2 , b 2 x 2 , cV e 2 .£ 2 , denote the first set of four 
 □'s ; aH 2 x 2 , b 2 t % x'\ c 2 t 2 x 2 , and e 2 t 2 x 2 , the four square numbers in (15). 
 Put a 2 — 4(^ 2 -|-</ 4 ), b 2 =2pq 2J r q 4: , c 2 =q 4 , ande*=£> 2 ; then must 
 4(P'/+?> 2 =D . . . [«], (2pq 2 J r q i )x 2 ={J • • • [ft], 
 
 [2] (jP+4j W , +4$*)flfeQ [7] 09 2 +6^ 2 + 9£> 2 =[J, 
 [4] (p+?>X<Z 2 ^-Q [0] (pV+W+4^)a^Q, 
 
 [5] (p-HF)*X(p+&*=*a> L 10 ] (ity+*ptf+f)**0, 
 
 [6] (6;>y 2 +9y*)aj 2 =n, [11] (9pY-\-18pq 6 +9q 8 )x < =\J. 
 
 These are all squares except [f#], [ft], [1], [4], [6], which reduce 
 
 □, and 6jj-\-9q i =\^\ ; [#«] and [4] being identical when reduced. 
 Since q 2 , p-{-q 2 , and 2/9-j-y 2 will be three square numbers in 
 arithmetical progression, we may take 
 
 q—r 2 — 2rs — s 2 and p=4rs(r 2 — s 2 ) ; then will 
 
 a 2 .r 2 =4(;- 2 — 2rs— « J ) a (j- 2 +* a )V=4(jpy a +^)aj 2 , 
 b*x 2 =(r*+2rs— s 2 )\r 2 — 2rs— s 2 ) 2 x 2 =Q2pq 2 +q i )jy 2 , 
 c 2 x 2 =(r 2 —2rs—s 2 Yx 2 =q i x\ 
 e 2 x 2 '=l Qr 2 s\r 2 —s 2 ) 2 x 2 =p 2 £ 2 . 
 Substituting the above values oip and q in [6], it becomes 
 9r*— 12r 3 s+18rV-|-12rs 3 -f 9s 4 ==Q say=(3r 2 -2rs-|-36' 2 ) 2 ; then *==6*. 
 . ' . 4(pq 2 +q i )x 2 =(\ , 702ys 8 x 2 , (2pq 2 ^ r q i )x 2 =(l08l') 2 s 8 x 2 , 
 q*x i =(2S') 4 ^x\ p 2 x 2 =( 840) V,« 2 . 
 To satisfy the first condition of the problem, assume 
 
 (a-\-b-\-c-\-e)x=m*\ then x=m 2 -^(a-\-b-\-c-\- e). 
 Premising that either the fractional or integral expressions in 
 the first four of the fourteen following lines are the squares in 
 [12]; the next six, those in [13]; the last four, those in [14]; 
 the second expression in each line being produced by multiplying 
 each term in the first expression by (a-\-b-{-c-\-e) 2 -T-m 2 , we have 
 
 (a+ ^ +e) , +ggffg, - a ^ + <<b + e + e) + « + c + ey, 
 
23 
 
 , — nn — r— w-H~ , A , — i — , or &*m*+£(a+c +«)+(«+ «-H)i 
 c 2 m* , («+*+<?)m a .. - 
 
 (tf+&+c+g) . +5+g+3+7 » or ^•+^+*+»>+(*+*+*yi 
 ( 444«) > + ^+fc^? or («+*>'" J -K«+«x c + e >-H c + e >'» 
 
 (a-A-eYm 4, , (£+6>fra J , ■ U * i / i \// i \ i rx i \t 
 
 ^V$i R ) 2 +^++ e . « («+e)W+(«+ C X*-N>-M6+«)'. 
 l££fifr$£&k « (« +e) W +( , + ,)(VH C ) + (H-or, 
 
 (■Sg^i ^ife - <* + e)W+<* + .)< tt+ ,) + <<H-<-)'. 
 (o^^+^tgi. « (.+«)•*•+(*+«)<« + *> + («+*)•, 
 
 (" + /> + C) 2 y//* , 0?» 8 • . - . \o 2 | / | / | X I ,2 
 
 \a + 6+c+er + a+h+c + e > ° r («+*+«) W +t»+»-M«+*. 
 
 (a+£+e)*m* . m 8 , , 7 , w» i / i x i \ i * 
 
 ; a+6 + c 4. e) , + ^ +ft+c+e , or (« + ft + e )W + (<, + S + e ) C -r«, 
 
 (" + '' + e) 2 m 4 , bur , . , \t--i i / i i \x i xj 
 
 ( a +6+4*f + «+6+c+e ' « U l + c + ehn- + ( a + c + e)0 + / l , 
 (b-\-c J t-eYm i ' am' /7 , - no 2 i // i i \ 2 
 
 ■ ( g +6+ C +e)' + a -H+c+ e '° r (*+«+*)'»» , +(*+«+<0«+« • 
 
 To find m, put a*m s +(d+<J+e)a+( x +c+€) 2 =(am+6+c+e) 8 ; 
 then will /»=<£•, and it is obvious that the same value of m will 
 fulfill all the remaining conditions. 
 
 y/r 1 1 
 
 a+d+C+6 4( 1 702 -r- 1081+529 + 840)tf 4 16608s 4 
 
 1702s 4 1702 7 1081 529 840 
 
 ax=^-zr — =— — , &«= — , cx= - , ese: 
 
 To fulfill the first tli ree conditions of (15), let e'=840, and write 
 1702 for a, 1081 for £, 529 for c, (1081+529+840) for </, (1702+529 
 + 840) for e, (1702+1081 + 840) for/, to for a; then must </ 2 ^; 2 -j- 
 </to, b''t' l x' 2 -\-etx, cH 2 x 2 -}-ftx, be Q's. Place the above values in for- 
 mula (14), ]J. 1H, put A 2 for the square numerator in the formula, v 
 for the prodnct of the three trinomial factors in the denom., and B, 
 C, D, E, for the denoins., respectively, of to, atx, btx, etx\ then will 
 
 frm A 2_i_o ,,2 J .2^2 „ ( 12 5 4 6 4 1) 9 4 2 4 2 7 8 5 2 8 G 4 1 9 2 1 9 2_4 M 7 . 
 
 2 5 4 6 4 9 9 4 2 4 2 7 8 528 6_4_1 B 21 9 2 4) 5 2 6 2 9 2 2 7 10 7 H-U5A6 8 2 9_ 
 
 •<T#9T 23 0T7 T6 o 5TTo 6).(1 5 7 tfrftffftl 2T8) — i O&SlS fff If 5 11.31 4 7^0 1 
 
 n g i o 7 93 Qoj a a 4 1 boj i i 9 5 1; 4 :; 2 9 2 3. c 7 9 9 9 0792 6 6j 7 7 6 — A 2 — U 
 
 8 7 5 1 3 6 9(1 9 4 M 7 2 6 6-1 9 (J 2 6 2 6 5 5 4 7 8 4 40 1 48 3 3 "3 ""» 0(1 8 - 1 ' °> 
 
 a 2 ^ 2 =rA 4 -^-(8«^ 2 c 2 ^) 2 =A 4 ~(589581 24486219416687353 1 22027580- 
 
 771248509165843419851 24795231104) 2 =A 4 H-C 2 =D. 
 /;%: 2 =A 4 ^-(8a 2 /3c 2 ?j) 2 =A 4 ^-(92827685361 28163478434321 3105127- 
 
 171752971878136451441 11379725568) 2 ^A 4 --D 2 =: [J, 
 c 2 ^ 2 =A 4 ^-(8a 2 /3 2 ct0 2: =A 4 -f-(189691357042618992820l7961000178- 
 
 161184302949010492251010210743552) 2 =A 4 -T-E 2 =Q, 
 ^ 2 ic 2 r=(AV) 2 H-B 2 =(44208550770666633364071066121 1206865140- 
 3864036558911912665835891840) 2 --B 2 =n, for the four 
 squares required to satisfy (15) ; since the values of ax, bx, ex, 
 ex, fulfill conditions (2, 3, 4, 5, 6, 7, 8, 9, 10, 11); </to, btx, ctx, etx, 
 will also satisfy the same conditions, and formula (14), p. 18, gives to 
 such a value that atx, &to, <fcc, will fulfill the other conditions of (15). 
 
24 
 
 £W+eta=(£) 2 (AM-8aVet>)> c 2 ^ 2 +/^=(^) 2 (A 2 +8a 2 ^); and 
 since each of these three binomial factors is a □, 
 . \ « 2 ^ 2 +^te==(^) 2 (A 2 -|-8Z» 2 cV^)=(^) 2 (ll7259729972169l775449- 
 94691400724) 2 =Q 
 ^V 2 +^^'=(^) 2 ('A 2 +8aV6v)=(ft) 2 (l7786013907038436904858l7- 
 
 32007492) 2 =Q 
 c 2 ^ 2 +/^==(^) 2 (A 2 +8aWA , )= : (E) 2 (367665873l41977567378862- 
 842430140) 2 =n. 
 
 24. To find n numbers, such that if the square of tlic first be 
 added to the second, the square of the second to the third, the 
 
 square of the third to the fourth, and the square of the nth. 
 
 to the first, the respective sums shall all be squares. 
 
 Solution. (1) For two numbers, x, y\ then x' iJ r y=[^}, an( l 
 y--\- .>•=□• The first expession is a square when y=2x-{-l, and the 
 
 second becomes 4ar J -j-5.r-|-l==n= ( P-x—l\ \ then x= 2pq ~^~ 5g2 . 
 
 (2) For three jSTos., ./', y, z; then x 2 -\-y=z\2, V 2 + z =0, z 2 -j r x=\J. 
 The first expression is a square when y=2x-{-l, the second when 
 
 2=4.r-f3, and the third becomes 16.r 2 +25«-f9==n]== f-^-x— s\\ 
 
 Reducing, x=z &Z T * % * (3) For four numbers, x, y 9 a, w; taking 
 y=2x-{-l, z=4x-\-3, w=Sx-\-1 9 the first three expressions will be 
 squares, and the fourth becomes 64x 2J r ll3x-\-49=£2= (£-x— 7 J 
 
 Reducing, x — I -—-_' and thence y, z, u\ are known. 
 
 p 2 — 64</" 
 
 (4) For n numbers, we perceive now the derivation of the suc- 
 cessive assumptions of y, z, w, &c, from that immediately pre- 
 ceding-, viz., multiplying the first term by 2, and the second by 2 
 and adding 1, as follows: CB+O, 2flJ+l, 4.2 -f-3, 8#+7, 16^+15, 
 32.r-j-31, &c, in each of which assumptions, the coefficient of the 
 first term is a power of 2, and that of the second, the same power 
 of 2 less 1. Now, putting x=number of quantities less 1, the last 
 assumption will be 2 x aj-J-2 K — l,=the value of the ?ith quantity, & 
 tlie square of this + ; r=2* x a5 2 +(2* x + 1 — 2 w + 1 -f-l)a;+(2 N — l) 2 =[j=: 
 
 (> ,_(2*_1) ) 2 ; then will ^ *P9t2*-W*™-*^ + »£ 9 
 and thence all the remaining quantities become known. 
 
 For three numbers, let x— 2, then x= 1 J Q~^ °9 agreeing with 
 
 p' 2 — lQq 2 ' ■ fo 
 
 the value of x in (2). For four Kos., let n=3, then X= 14 P2+ US <1* 
 J ' p 2 — 64? ' 
 
 agreeing with the value of x in (3), etc.; in which values of x, p, 
 q, may be any numbers that will make the denominator positive. 
 
 By putting N=l, 2, 3, 4, <fcc, successively, each supposition will 
 give a different value of x\ thus in fact constituting answers to as 
 many distinct problems, each perfectly general, and from which 
 the preceding quantities, ?/, 2, w, <fcc, are readily found. 
 
This book is DUE °°______ T __ i — 
 
 R EC'D Ui 
 
 HAR2'0^ 7 
 
 LI 
 
 NOV 20 1949 ^gfflay'OT* 
 
 nov* ^ 54 I ' cw - MAY22 198 i 
 
 LD 21-10 Ow- 9 
 
 >48(B399sl6)476