, 2010

Suppose there are four such numbers. Without loss of generality, we may (and will) assume that they are jointly coprime: $(w, x, y, z) = 1$.

Reduction modulo $4$ shows that $z$ and exactly one of the numbers $w, x, y$, say $y$, must be odd.

Write
$$w^{2p} + x^{2p} = z^{2p} - y^{2p} = (z^2 - y^2) \left(\sum_{k=1}^{p-1} z^{2(p-k-1)} (z^{2k} - y^{2k}) - p z^{2(p-1)}\right)$$
to deduce that the second factor above is congruent to $3$ modulo $4$ and infer thereby that in its decomposition into prime factors some prime $q \equiv 3 \pmod{4}$ occurs with an odd exponent.

Since $-1$ is a quadratic non-residue modulo $q$, it follows that $w$ and $x$ are both divisible by $q$, and in the decomposition of $w^{2p} + x^{2p}$ into prime factors, $q$ occurs with an even exponent.

Hence $z^2 - y^2$ is divisible by $q$, and therefore so is $p z^{2(p-1)}$. Notice that $p \neq q$ (for $q$ divides $w$, but $p$ does not by assumption) to deduce that $z$ is divisible by $q$. Then so is $y$.

Consequently, $w, x, y, z$ all share the common factor $q$, in contradiction with their joint coprimality.