534 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xx 



Euler 5 applied .to z 3 +l = D his 144 method of Ch. XXII to make a cubic 

 or a quartic a square, finding no solutions except 0, 1, 2, and stated that 

 there are no others. Cf. Euler 157 of Ch. XXI. 



Euler, 6 to make ax 2 +cy 2 a cube, assumed that 



x ^a+y V - c = (p Va + q V^c) 3 , 



whence x = ap 3 Scpq 2 , y = 3ap' 2 qcq 3 . For Fermat's case z 2 +2, we have 

 (Art. 193) a = 1, c = 2, y= 1, whence q(3p 2 2g 2 ) = 1, and q divides unity. 

 Taking q = l, we have 3p 2 2 = 1, whence p 2 = l, 2 = 25. A like proof is 

 given (Art. 192) of Fermat's result that 4 and 121 are the only integral 

 squares which when increased by 4 give a cube. But (Arts. 195-6) for 

 2x 2 5 the method leads to no solution, whereas the solution 2 = 4 exists 

 and the above assumption is shown to fail. 



A. M. Legendre 7 treated Fermat's problems as had Euler. 6 

 V. A. Lebesgue 8 proved that x 1 7 = y* is impossible. For, if y is even, 

 x is odd and z 2 = 8n+l =K2y) 3 +7; while, if y is odd, x 2 +l = (y+2)Q is 

 impossible since the prime divisors of Q = (y 1) 2 +3 are of the form 4n+3. 

 L. Ottinger 9 noted that x 2 ?/ 2 = Tz 3 has the general solution 



{4?n 3 3mr(2mr) } 2 - { (rar)(4m 2 d=2mr+r 2 ) } 2 = =F(2mrr 2 ) 3 . 



T. Pepin 10 criticized Euler's 6 proofs, noting that there may exist sets 

 of formulas for x and y other than the set deduced by Euler's assumption. 

 He proved Fermat's 2 ' 3 assertions. He studied the solution of x z -}-cn 2a = z 3 

 for c = 1, 2, 3, 4, 7, n being 1 or an odd prime, and z being odd if c = 7, and 

 proved that the following are not cubes: z 2 -f-l (z>0); 2 +3; 4or+7; 

 z 2 +9; x 2 +n 2 if n = 10Sl+k(k = 23, 35, 59, 71, 95), or n = 83, 263, 407, or if 

 n is a prime 12Z+7 with 7<n<1350; x 2 +2n 2 if n is a prime 24Z+5 or 

 24Z+7; 2 +3n 2 if n is a prime 6/+5 or its square; x 2 +5. Also, x 2 +9 2 = 2 3 

 only for x = 46, z 2 +7 2 = 2 3 only for z = 524; z 2 +H 2 = z 3 only for z 2 = 4. 



H. Brocard 11 and others gave various solutions of x 3 +17 = ?/ 2 . 



G. C. Gerono 12 proved that x s = y 2 +17 is impossible in integers by use of 

 (a; +2) {(x 1) 2 +3} =?/ 2 +5 2 and the divisors of a sum of two squares. 



E. de Jonquieres 13 proved that x 3 +a = y'* is impossible in integers if 

 a = c 3 -4, \c =1, 3, 7 (mod 8), or o = c 3 -4', c | = 3, 5, 7 (mod 8), t>l, or 

 a = c 3 -!, c = 2(2d+l), and hence if a= -3, -5, 7, -9, 11, -17, 23, -43, 

 61; also for a = 4, 6, 14, 16 if o;+0. 



F. Proth 14 stated and E. Lucas 14 proved that .T 2 +3 = ?/ 3 is impossible 

 since 7/ = r 2 +3s 2 , while x 2 3 = ?/ 3 holds only for x = 2, y = l. 



5 Algebra, 2, 1770, Ch. 8, Art. 121; French transl., 2, 1774, pp. 135-152; Opera Omnia, (1), 



I, 392. M6m. Acad. Sc. St. Petersbourg, 11, 1830 (17SO), 69; Comm. Arith. Coll., 

 II, 478. 



6 Algebra. II, Ch. 12, Arts. 187-196. Opera Omnia, (1), I, 429-434. 



7 Theorie des nombres, ed. 3, II, 1830, Art. 336, p. 12. 



8 Nouv. Ann. Math., (2), 8, 1869, 452-6, 559. 



9 Archiv Math. Phys., 49, 1869, 211. 



10 Jour, de Math., (3), 1, 1875, 318-9, 345-358. Details in Pepin. 72 



11 Nouv. Corresp. Math., 3, 1877, 25, 49; 4, 1878, 50. Cf. Escott, 37 Brocard. 65 



12 Nouv. Ann. Math., (2), 16, 1877, 325-6. 

 Ibid., (2), 17, 1878, 374-380, 514-5. 



" Nouv. Corresp. Math., 4, 1878, 121, 224. 



