Some interesting characteristics of any putative counter-example of
Fermat's Last Theorem can be deduced quite simply, and can be used
to place very strong constraints on the set of possible counter-
examples. One of these was noted by Barlow in 1810 (and probably
by others earlier). Assume non-zero integers x,y,z such that
x^p + y^p + z^p = 0 (1)
Without loss of generality we consider only primitive solutions, i.e.,
those for which x,y,z are pairwise coprime. Now we wish to show that
the quantity (x+y+z)^p is of the form pABCD where A,B,C,D are pairwise
coprime integers with
A = x+y B = x+z C = y+z
To prove this, notice that the binomial expansion allows us to write
[(x+y) + z]^p = (x+y)^p + p (x+y)^(p-1) z
+ [p(p-1)/2] (x+y)^(p-2) z^2
...
+ p (x+y) z^(p-1)
+ z^p (2)
Also, since
-z^p = x^p + y^p
= (x+y) [x^(p-1) - x^(p-2) y + ... + y^(p-1)] (3)
we know the last term is divisible by (x+y), and so the whole quantity
[(x+y)+z]^p is divisible by (x+y). By symmetrical arguments, we can
show that this quantity is also divisible by (x+z) and (y+z). Further,
it's clear that (x+y), (x+z), and (y+z) must be pairwise coprime,
because (for example) any prime that divides (x+y) also divides
(x+y+z) as we've just seen, and so it must divide z. It follows that
it a given prime divides TWO of the numbers (x+y), (x+z), (y+z), then
it divides two of the numbers z,y,x, contradicting the pairwise
co-primeness of these these basic numbers.
Equation (3) also shows that the integer (z^p)/(x+y) is coprime to
(x+y), except for a possible factor of p. To see this, suppose the
contrary, i.e., that the right hand factor of (3) is congruent to zero
modulo some prime divisor q of (x+y). This means x=-y (mod q), so we
can substitute -x for y into the right hand factor of (3) to give the
putative congruence
p x^(p-1) = 0 (mod q)
Obviously since q divides x+y it cannot be a divisor of x (recallng
that x and y are mutually coprime), so this congruence is impossible
unless the prime q happens to equal the exponent p.
Therefore, once we have divided equation (2) by (x+y), including taking
a factor of (x+y) out of the last term z^p, each of the remaining terms
still has a factor of (x+y) or z (which shares every prime divisor of
(x+y)), with the exception of the last term, z^p/(x+y), which we have
just seen is coprime to (x+y), with the possible exception of a factor
of p. We can also see from the binomial coefficients of (2), excluding
the terms x^p + y^p + z^p which vanish, every one is a multiple of p,
so the entire quantity (x+y+z)^p is a multiple of p, as well as being
a multiple of the mutually coprime quantities A=(x+y), B=(x+z), and
C=(y+z), at most one of which is itself a multiple of p. Thus, we
have
(x+y+z)^p = pABCD
where A,B,C,D are pairwise coprime integers. Since the left-hand
side is a pth power, and the right-hand factors are coprime, it
follows from unique factorization that each of the right-hand factors
is a pth power power, except for whichever one (if any) of them is
a multiple of p, which must be p^(p-1) times a pth power. Hence if
none of the integers x,y,z is divisible by p (i.e., in Case 1 of
Fermat's Last Theorem) we have integers a,b,c,d such that
x+y = a^p x+z = b^p y+z = c^p pD = d^p
On the other hand, if one of the base variables, say, z, is divisible
by p (i.e., Case 2), then we have
p(x+y) = a^p x+z = b^p y+z = c^p D = d^p
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