2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ An obvious first thing to do is to clear denominators and write the equation as $(x^4 + y^4) z^2 = x^4 y^4$. $\endgroup$ – Michael Joyce Apr 18 '13 at 13:29
  • 1
    $\begingroup$ @Benzne_O between Michael's comment and the theorem you specified, you've solved the problem.. can you see it? $\endgroup$ – Vincent Tjeng Apr 18 '13 at 14:36
2
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.