# Diophantine Equations (Reciprocals?)

I'm having difficulty solving the following problem:

Consider the following Diophantine equation

$$\frac{1}{x^4} + \frac{1}{y^4} = \frac{1}{z^2}$$

Show that this equation has no solutions, where $x, y, z \in \mathbb{Z}^+$

I know there is a theorem that says that $x^4 + y^4 = z^2$ has no solutions, but I'm not sure how to connect that with its reciprocals. I'd appreciate any help.

• An obvious first thing to do is to clear denominators and write the equation as $(x^4 + y^4) z^2 = x^4 y^4$. – Michael Joyce Apr 18 '13 at 13:29
• @Benzne_O between Michael's comment and the theorem you specified, you've solved the problem.. can you see it? – Vincent Tjeng Apr 18 '13 at 14:36

## 1 Answer

First multiply both sides by $x^4 y^4$ which generates the following: $$\frac{x^4 y^4}{x^4} + \frac{x^4 y^4}{y^4} = \frac{x^4 y^4}{z^2}$$

$$y^4 +x^4 = (\frac{x^2 y^2}{z})^2$$

Which is obviously a contradiction because of the fact that $x^4 + y^4 = z^2$ has no solution (ie if the above equation held, then by simple algebraic manipulation you can deduce a fact that is known to be false)