Fermat’s Last Theorem extends to negative Integers

I bumped into Fermat’s Last Theorem the other day, probably the most famous margin note in history:

$a^n+b^n=c^n$

Where $a$ , $b$ , $c$ and $n$ are all positive integers. The equation has no solutions for $n>2$ .

But the theorem always constrains $a$ , $b$ and $c$ to be positive. Why?

Here’s what I found plugging in $-a$ and $-b$ scenarios into the formula.

First, if $n$ is even, $(-a)^n=a^n$ so this returns us to the positive Integer version of Fermat’s Last Theorem.

Conversely, if $n$ is odd and we have both $-a$ and $-b$ , then:

$(-a)^{n_{odd}}+(-b)^{n_{odd}}=-(a^{n_{odd}}+b^{n_{odd}})$

This is just the negative version of Fermat’s Last Theorem, so it follows again there’s no solution for $n>2$ .

Next, I discovered an infinite number of solutions for the equation to any odd power. If we set $b=-a$ , this leads to:

$a^{n_{odd}}+(-a)^{n_{odd}}=a^{n_{odd}}-a^{n_{odd}}=0^{n_{odd}}$

Let’s just declare that solution “trivial” and move on. So the final leftover scenario is this:

$a^{n_{odd}}+(-b)^{n_{odd}}=c^{n_{odd}}$

Which can be rearranged to this:

$a^{n_{odd}}=b^{n_{odd}}+c^{n_{odd}}$

And this, of course, is just another rewrite of Fermat’s Last Theorem.

So throwing away the trivial case that $b=-a$ with $n_{odd}$ it quickly becomes clear that the theorem is true for all negative numbers with $n>2$ . Which still makes me wonder why the theorem is always stated “for positive $a$ , $b$ and $c$ “?

PS. When I was initially googling to find other work on this question nothing relevant popped up (leaving me to figure out how negative numbers work in the theorem myself). However, I found other interesting proofs. Here’s a simple proof extending Fermat’s Last Theorem to all positive rationals $a$ and $b$ :

Write Fermat’s Last Theorem with rationals:

$(\frac{a}{x})^n+(\frac{b}{y})^n=(\frac{c}{z})^n$