Find all integer solutions to $x^2+y^2+z^2=2xyz$

Question

I am working on some of these types of problems for fun, just want to see a couple solved as examples.

Answer 1

From $z^2 - 2 (xy) z + (x^2 + y^2) = 0$, we get $$z = xy \pm \sqrt{x^2 y^2 - x^2 - y^2}$$ so we basically need to search for $x, y$ such that $x^2 y^2 - x^2 - y^2$ is a square say $$x^2 y^2 - x^2 - y^2 = t^2$$ or $$(x^2 - 1) (y^2 - 1) = t^2 + 1$$

Now $a^2 \equiv 0, 1 \mod 4$ hence $a^2 - 1 \equiv -1, 0 \mod 4$ corresponding to $a$ even, odd respectively so $LHS \equiv 0, 1 \mod 4$. But $t^2 + 1 \equiv 1, 2 \mod 4$. So we must have $x, y, t$ are all even. Write $x = 2x_1$, $y = 2y_1$ and $t = 2t_1$ we get $$16 x_1^2 y_1^2 - 4x_1^2 - 4y_1^2 = 4t_1^2$$ which reduces to $$4 x_1^2 y_1^2 - x_1^2 - y_1^2 = t_1^2$$ Now $x_1^2 + y_1^2 + t_1^2 \equiv 0 \mod 4$ and this forces $x_1, y_1, t_1$ be all even. Write $(x_1, y_1, t_1) = 2 (x_2, y_2, t_2)$ and we once again has $$4 \cdot 16 x_2^2 y_2^2 - 4 x_2^2 - 4 y_2^2 = 4 t_2^2$$ which reduces to $$16 x_2^2 y_2^2 - x_2^2 - y_2^2 = t_2^2$$ and we again deduce $x_2, y_2, t_2$ are all even. This argument can be repeated ad-infinitum which means $x, y, t$ must be divisible by arbitrary power of $2$ and that can only happen when $x = y = t = 0$. So the only solution to your equation is $(x, y, z) = (0, 0, 0)$.

Answer 2

First we know $x^2+y^2+z^2$ is even so either two odd one even or three even.

In the case of two odd one even we have $x^2+y^2+z^2\equiv2\pmod{4}$ while $2xyz\equiv0\pmod{4}$ so this is impossible so all three are even.

Let $x=2a,y=2b,z=2c$ we have $a^2+b^2+c^2=4abc$. Now $a,b,c$ must be all even again.

Let $a=2a_2,b=2b_2,c=2c_2$ we have $a_2,b_2,c_2$ must be all even again as the right side will again be a multiple of four.

Hence by infinite descent we can show there is no solution other than $(0,0,0)$, which is the only triple that can afford infinite divisions by two and still be even.