Why z Is Not a Prime Power in z^p = x^p + y^p

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 x^p + y^p = z^p 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 x^2 + y^2 = z^2 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
x^p + y^p = z^p 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 x^p + y^p = z^p 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 out the simple proof that x^p + y^p = z^p
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

             (p-1)/2            (p-j-1)!
 x^p + y^p  =  SUM   (-1)^j  p ---------- (x+y)^(p-2j) (xy)^j      (1)
               j=0             j! (p-2j)!


The coefficient of the jth term (up to sign) is the sum of binomial 
coefficients
                     / p-j \       / p-j-1 \
                    (       )  +  (         )
                     \  j  /       \  j-1  /

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

 x^11 + y^11  =  (x+y)^11  -  11 (x+y)^9 (xy)  +  44 (x+y)^7 (xy)^2

                         -  77 (x+y)^5 (xy)^3  +  55 (x+y)^3 (xy)^4

                                               -  11 (x+y) (xy)^5

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

                  x^p + y^p  =  (q^m)^p

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) = q^s, and it's clear that s must be greater than 0, because
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) = (x^p + y^p), which is impossible in integers greater than 1.

So, settiing equation (1) equal to q^(mp) and dividing by (x+y) = q^s,
we have

  (p-1)/2            (p-j-1)!
    SUM   (-1)^j  p ---------- (q^s)^(p-2j-1) (xy)^j   =   q^(mp-s)
    j=0             j! (p-2j)!

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

                         p (xy)^((p-1)/2)

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

                 x^p + y^p
                 ---------  =  p
                   x + y

However, if p is an odd prime and both x and y exceed 1, we have x^p 
greater than px, and y^p greater than py, which implies that (x^p+y^p)
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 2^p = 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 x^p + y^p = z^p 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.

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