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[dire wraith]

i found that x^2 +y^2 + w^2 is congruent to z^2 module p

idk what to do after that

[fast lagoon]

WLOG assume $\gcd{(w, x, y, z)} = 1$.
If $z$ is even, taking modulo $4$ gives $w, x, y$ are also even, contradicting our assumption.
So $z$ is odd. So at least one of $w, x, y$ is odd. WLOG let $y$ be odd.
We have $w^{2p} + x^{2p} = z^{2p} - y^{2p} = (z^2 - y^2)((z^2)^{p-1} + (z^2)^{p-2}y^2 + \dots + (y^2)^{p-1})$.
Let $T = (z^2)^{p-1} + (z^2)^{p-2}y^2 + \dots + (y^2)^{p-1}$.
Now since $T \equiv 3 \pmod{4}$, there exist a prime $q \equiv 3 \pmod{4}$ such that $v_q(T)$ is odd.
But since $w^{2p} + x^{2p}$ is a sum of two squares, we have that $v_q(w^{2p} + x^{2p})$ is even, and so $v_q(z^2 - y^2)$ is odd.
So we have $q \mid z^2 - y^2$ and $q \mid T$, implying that $q \mid pz^{2(p-1)}$.
If $q \neq p$, we would have $q \mid z \implies q \mid y$. And since $q \mid w^{2p} + x^{2p} \implies q \mid w$ and $q \mid x$, again we contradicted our assumption that $\gcd{(w, x, y, z)} = 1$.
So $q = p$. And so we have $w^{2p} + x^{2p} \equiv 0 \pmod{p}$. If $p \nmid w, x$, we would have $\left(\frac{w}{x}\right)^2 \equiv -1 \pmod{p}$, but this is a contradiction since $-1$ is not a quadratic residue modulo $p$. And so we must have $p \mid w$ and $p \mid x$ $\box$

[soft thistle]

trigonometria
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