Exercise 5.4.4 (Equation $x^2 + y^2 + z^2 = 2xyz$)

Proof. Assume for the sake of contradiction that there are integers $x,y,z$ such that $x^2+y^2+z^2=2\mathit{xyz}$, where $(x,y,z)\ne (0,0,0)$. Since $x,y,z$ are not all equal to $0$, there exists a largest integer $k$ such that $2^k\mid x,2^k\mid y,2^k\mid z$. Then there are integers $u,v,w$ such that

and $u,v$ or $w$ is odd (otherwise $2^{k+1}\mid x,2^{k+1}\mid x,2^{k+1}\mid x$, in contradiction with the definition of $k$). Then $2^{2k}(u^2+v^2+w^2)=2^{3k+1}\mathit{uvw}$, and after simplification by $2^{2k}$,

Without loss of generality, we can suppose that $u$ is odd, since the other cases $v$ odd, $w$ odd are similar. Since $2^{k+1}\mathit{uvw}$ is always even, $v^2+w^2$ is odd, thus $v$ and $w$ have not same parity, so we can assume $v$ odd and $w$ even. Then $4\mid 2\mathit{uvw}$, thus

But $u^2\equiv 1,v^2\equiv 1,w^2\equiv 0(\mathit{mod}4)$, thus $u^2+v^2+w^2\equiv 2(\mathit{mod}4)$, and we obtain $2\equiv u^2+v^2+w^2\equiv 0(\mathit{mod}4)$, which is impossible.

This contradiction shows that there don’t exist integers $x,y,z$ such that $x^2+y^2+z^2=2\mathit{xyz}$, where $(x,y,z)\ne (0,0,0)$.

If $x,y,z$ are integers such that $x^2+y^2+z^2=2\mathit{xyz}$, then $x=y=z=0$. □