this is in pftb? cites it as an old amm problem as well

@above yeah it's in page 81. I will post the solution from there. Without loss of generality we can assume $\gcd{(x,y,z,t)} = 1$. If $t$ is even, taking modulo $4$ gives $x,y,z$ are also even, contradicting our assumption. So $t$ is odd. So at least one of $x,y,z$ is odd. WLOG let $z$ be odd. We have $x^{2p} + y^{2p} = t^{2p} - z^{2p} = (t^2 - z^2)((t^2)^{p-1} + (t^2)^{p-2}z^2 + \dots + (z^2)^{p-1})$. Let $T = (t^2)^{p-1} + (t^2)^{p-2}z^2 + \dots + (z^2)^{p-1}$. 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 $x^{2p} + y^{2p}$ is a sum of two squares, we have that $v_q(x^{2p} + y^{2p})$ is even, and so $v_q(t^2 - z^2)$ is odd. So we have $q\mid t^2 - z^2$ and $q\mid T$, implying that $q\mid pt^{2(p-1)}$. If $q\neq p$, we would have $q\mid t \implies q\mid z$. And since $q\mid x^{2p} + y^{2p} \implies q\mid x$ and $q\mid y$, again we contradicted our assumption that $\gcd{(x,y,z,t)} = 1$. So $q=p$. And so we have $x^{2p} + y^{2p} \equiv 0 \pmod{p}$. If $p\nmid x,y$, we would have $(\frac{x}{y})^2 \equiv -1 \pmod{p}$, but this is a contradiction since $-1$ is not a quadratic residue modulo $p$. And so we must have $p\mid x$ and $p\mid y$, as desired.

khina wrote: this is in pftb? cites it as an old amm problem as well What is pftb

PNT wrote: khina wrote: this is in pftb? cites it as an old amm problem as well What is pftb problem from the book of titu andresscu