# Prove that the diophantine equation $1/x^4 + 1/y^4 = 1/z^2$ has no solutions in nonzero integers, x, y, z

Prove that the Diophantine equation $1/x^4 + 1/y^4 = 1/z^2$ has no solutions in nonzero integers, x, y, z.

I've tried to solve this using the Theorem: "The Diophantine equation $x^4 + y^4 = z^2$ has no solutions in nonzero integers x, y, z.", and transformed the equation into $y^4z^2 + x^4z^2 = x^4y^4$. What are the next steps?

The next step is: $z^2(x^4+y^4) = x^4y^4 = (x^2y^2)^2\implies x^4+y^4 = k^2$, contradiction to the result you quoted.
• I follow that $(x^2y^2)^2$ is $k^2$, but that means either one of two things: 1. $z^2(x^4 + y^4)$ is equivalent to $x^4 + y^4$ which means that $z^2x^4$ has an integer fourth square (same for z^2y^4) or 2. $k^2 / z^2$ is a perfect square but I don't see how? – Patrick Pei Dec 13 '16 at 9:33