CHAPTER XXVI. 



FERMAT'S LAST THEOREM, ax r +by s = cz t , AND THE CON- 

 GRUENCE x n +y n =z n (MOD p}* 



For proofs of the impossibility of x n +y n =z n for n = 3, 4, see Chs. 

 XXI, XXII. 



Leo Hebreus, 1 or Lewi ben Gerson (1288-1344), proved that 3 m l 4=2" 

 if m>2, by showing that 3 OT d=l has an odd prime factor. The problem 

 had been proposed to him by Philipp von Vitry in the following form : All 

 powers of 2 and 3 differ by more than unity except the pairs 1 and 2, 2 and 3, 

 3 and 4, 8 and 9. 



Fermat, 2 commenting about 1637 on Diophantus II, 8 (to solve 

 x 2 +y 2 = a 2 ), stated that "it is impossible to separate a cube into two cubes, 

 or a biquadrate into two biquadrates, or in general any power higher than 

 the second into two powers of like degree ; I have discovered a truly remark- 

 able proof which this margin is too small to contain." This theorem is 

 known as Fermat's last theorem. 



Claude Jaquemet 3 (1651-1729), in a manuscript in the Bibliothe'que 

 Nationale de Paris and first attributed to Nicolas Malebranche 4 (1638-1715), 

 attempted to prove Fermat's last theorem. In a z =x z +y z we may suppose 

 x, y relatively prime. The quotient of x z +y z by x+y is 



Then x-\-y and Q have no common divisor d other than factors of z. For, it 

 would divide 



Q- (x'-i+yx*- 2 ') = -2yx z ~ 2 +y 2 x z - 3 ---- y*~ l . 



Adding 2yx z ~ 2 +2y 2 x z ~ 3 , we get 3y 2 x z ~ 3 ---- . Finally, we get zy*~ l . But 

 y is not divisible by d since x, y are relatively prime; hence z is. Similarly, 

 x y and (x z y z }l(xy) have no common divisor not a factor of z. 



Suppose that a, x, y are relatively prime integers for which a z =x z -\-y z t 

 z odd. As just proved, at most one of the powers is divisible by z. First 

 let x* and y z be not divisible by z. Let x z = p z q z , y z = r z s z , where r and s 

 are. relatively prime, also p and q. Then apq = r z , ars = p z . Thus the 

 divisor p r of p z r z divides pqrs. Dividing the latter by p r, we get 

 the remainder pqps or rqrs, neither zero, and " by continuing this 

 process to infinity, we get no new remainders, so that pr is not a divisor 

 of pqrs." As pointed out by E. Lucas 40 the last conclusion is wrong; 



* H. S. Vandiver read critically the proof-sheets of this chapter and believes that the reports 



are accurate. Both he and the author compared the reports with the original papers 



when available. 



1 Cf. J. Carlebach, Diss. Heidelberg, Berlin, 1909, 62-4. 

 2 Oeuvres, I, 291; French transl., Ill, 241. Diophanti Alexandrini Arith. libri sex, ed., S. 



Fermat, Tolosae, 1670, 61. Precis des Oeuvres math, de P. Fermat, par E. Brassinne, 



Mem. Acad. Sc. Toulouse, (4), 3, 1853, 53. 

 1 Cf. A. Marre, Bull. Bibl. Storia Sc. Mat. Fis., 12, 1879, 886-894. 



* Cf. C. Henry, ibid., 565-8. 



40 Ibid., 568. Since he omitted the factor p before q s, take k to be a multiple of p. 



731 



