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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