Let $x, y, z, t$ be natural pairwise relatively prime numbers such that
$$xy+yz+zt = xt$$
Prove that the sum of squares of two of them is exactly twise the sum of squares of the other two numbers.
I figure that $x^2 + t^2 = 2(y^2 + z^2)$, but I can't prove it.
I've gone through tonnes of algebra only to figure out something else is needed. I could't even find a way to utilize the fact the numbers are relatively prime, as Bezout's lemma makes the algebra even messier.