# Fermat's Last Theorem related question

1. Feb 26, 2013

### Ninty64

1. The problem statement, all variables and given/known data
Show that $x^{n}+y^{n}=z^{n}$ has a nontrivial solution if and only if the equation $\frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}}$ has a nontrivial solution.

2. Relevant equations
By nontrivial solutions, it is implied that they are integer solutions.

3. The attempt at a solution
I was able to solve in one direction
Given an integer solution to $\frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}}$
Then it follows that:
$\frac{x^{n}+y^{n}}{x^{n}y^{n}}=\frac{1}{z^{n}}$
$x^{n}y^{n}=z^{n}x^{n}+z^{n}y^{n}$
$(xy)^{n}=(zx)^{n}+(zy)^{n}$
Therefore, since the set of integers is closed under multiplication, then $x^{n}+y^{n}=z^{n}$ has a nontrivial solution.

However, I can't seem to prove the other direction. Working the above proof backwards doesn't work unless I assume that d = gcd(x,y) and somehow prove that z = xy.

2. Feb 26, 2013

### haruspex

Applying exactly the same method in the reverse direction appears to work.

3. Feb 26, 2013

### hapefish

I might note that $\frac{1}{\frac{1}{x^n}}+\frac{1}{\frac{1}{y^n}}=\frac{1}{\frac{1}{z^n}}$ is identical to writing $x^n+y^n=z^n$. Then you can refer back to your earlier proof.

4. Feb 26, 2013

### tiny-tim

alternatively, since the formula you start with is symmetric, try doing something symmetric to it