Why z Is Not a Prime Power in zp = xp + yp

 

One of the fascinating things about Fermat's Last Theorem is that, although the general case is extraordinarily difficult, it is possible to dispose of many special cases with quite elementary arguments, as shown by Sophie Germain in her treatment of Case 1. By equally elementary arguments we can show that xp + yp = zp can have no solution in positive integers if z is a prime, or any power of a prime. This is admittedly a severe restriction, but it's interesting to note that the primitive integer solutions of x + y = z can be decomposed (trivially) into solutions where z is a prime, and that the primitive integer solutions of x2 + y2 = z2 can be decomposed into solutions where z is a prime of the form 4k+1, whereas our proposition rules out the possibility of decomposing a solution of xp + yp = zp into solutions with prime z for any p greater than 2. It's conceivable that Fermat in (say) 1638 might have momentarily assumed that any solution with composite z must be factorable into solutions with prime z, and since the latter cannot exist, he might have (rashly) concluded that neither can the former. Of course, it's not the case that solutions of xp + yp = zp with composite z must be factorable, so we can't rule them out based on this argument, but it nevertheless shows that the set of solutions for exponent greater than 2 could not possibly be anything like the well-behaved families of solutions for exponents 1 and 2, but would necessarily be "sporadic".

 

So, it's worthwhile to sketch the simple proof that xp + yp = zp with p a prime greater than 2 can have no primitive solutions in positive integers if z is a prime, or even any power of a prime. We begin by noting the algebraic identity for any positive integers x,y and odd prime p

 

 

The coefficient of the jth term (up to sign) is the sum of binomial coefficients

 

 

and since j and p−2j are less than p for j greater than 0, it's clear that this coefficient is a multiple of p for all terms except j=0, for which it is 1. For example, with p=11 we have

 

 

Now, if we assume positive integers x,y,m and odd prime q such that

 

 

then since the left side is divisible by (x+y) the right side must be also, which implies that (x+y) is a power of q. Thus we have (x+y) = qs, and it's clear that s must be greater than 0, because x+y is greater than 1 (since both x and y are positive integers). Also, s cannot equal mp, because that would imply (x+y) = (xp + yp), which is impossible in integers greater than 1.

 

So, settiing equation (1) equal to qmp and dividing by (x+y) = qs, we have

 

 

Every term on the left side contains multiple factors of q, with the exception of the final term, which is (up to sign) of the form

 

 

so this term must be divisible by at least one power of q. If q does not equal p, then q must divide xy, which is impossible because we assumed x,y coprime to z. Also, if s is less than mp−1 the above term must be divisible by more than one power of q, so it is still necessary for q to divide xy, contradicting the assumptions.

 

On the other hand, if q=p there remains the possibility that s equals mp−1, in which case we have

 

 

However, if p is an odd prime and both x and y exceed 1, we have xp greater than px, and yp greater than py, which implies that (xp + yp) is strictly greater than p(x+y), ruling out the above equality. The only remaining possibility in distinct integers is x=1, y=2, but this requires 2p = 3p−1, which is not satisfied for any prime p greater than 3, and the case p=3, x=1, y=2 can be ruled out by inspection.

 

As a historical aside, the great Niels Abel once remarked that Fermat's equation xp + yp = zp with p an odd prime is impossible if any of the numbers x, y, z are prime powers (and also if z+x, z+y, or x−y is a prime). Needless to say, the preceding does not constitute a proof of Abel's statement because it covers only the number z (with the stipulation that x, y, and z are positive integers). The analogous statements for x and y are much more difficult, because we can't appeal to simple numerical inequalities to limit the range of possible solutions.

 

Return to MathPages Main Menu