ON THE THEORY OF NUMBERS. 151 



We may suppose that X is greater than 3, and that no two of the numbers 

 x, y, z admit any common divisor. And first, let none of them be divisible 



a \ i 



by i— a, a. still representing a root of the equation — =0. Since for x 



we may write a? x, we may assume that x, y, z are of the form 



x=a + (l—ayX, 

 y=b + (l- a yY, 

 z—c + (l—a.yZ, 



a, b, c denoting integral numbers prime to X, which evidently satisfy the con- 

 gruence a+6+c = 0, mod X. The equation x K +y K +z K =0 may then be 

 written thus 



(x+ay) (x + a 2 y) (x + a 3 y) (x-\rOt K ~ l y) — — z\ 



No two of the factors of which the left hand member is composed can have 

 any common divisor ; each of thein is therefore the product of a perfect \th 

 power by a unit; so that we may write, x+a s y=a? e(a.)v\ e(a) denoting 

 a real unit. Since v K is an actual number, it follows (remembering that X is 

 not an exceptional prime) that v is also actual ; hence v K is congruous, mod X, 

 to a certain integral number m. Eliminating m X e(a) between the two con- 

 gruences x+a? y=ma. p e(a), and x+a,~ s y=ma.~ p e(a), mod X, we find 

 a~ p (x+a s y) — x p (x+a~ s y)=0, mod X. For the modulus (1— a) this 

 congruence is identically satisfied *. That it should be satisfied, mod (1 — a) s , 

 we must have the relation (a + b)p=bs, mod X ; whence, putting 



s=£, mod X, 



a + b 



we have p=ks, mod X. Substituting this value for p, we find that the con- 

 gruence 



a-* s (>+a s y)-a* s (> + a- s y )=() 



is identically satisfied, mod (1— a) 3 ; but in order that it should be satisfied, 

 mod (1—a.y, we have the condition 



s 3 b(2k-l)(k-l)-3s(k-L .y"+kx")~0, mod X, 



where x" and y" are the values (fora=l) of the second derived functions 

 of a: and y with respect to a. This conditional congruence must be satisfied 

 for every value of s ; either therefore AseeI, mod X, or 2£=1, mod X. The 

 supposition £ = 1 is inadmissible; for it implies that a=0, mod X, contrary 

 to the hypothesis. Hence we must have 2A = 1, and a=b, or, by parity of 

 reasoning, a=b^^c, mod X. But also a+Z>+c=0, mod X, whence we again 

 infer the inadmissible conclusion «=6=c=0, mod X. 



in ordinary integral numbers, but also in any complex integers composed of Xth roots of 

 unity. The demonstration does not possess the same generality when X is an exceptional 

 prime satisfying the three conditions cited iu the text. In this case M. Kummer has only 

 shown that the equation ^+^+2^=0 is irresoluble when we suppose that x, y, z are 

 ordinary integral numbers prime to X, or else complex numbers containing the binary periods 

 a-\-a~ ', one of which has a common divisor with X. 



* Since X is divisible by (l-«) x_1 , and since <p{ a )=<p{\)+{ a .-\)<j,'{\)J r {x-\f'^S^. 



+ • • • , it is readily seen that, if r<X— 1, the conditions for the divisibility of <p(a) by 



(1-*)'' are 0(1)^0, 0'(1)=O, ^( r -^(l)=0, mod X. 



