Is $x = 2$ is the only real solution for $a^x + b^x = c^x$ when $(a,b,c)$ is a pythagorean triplet?

Question

Take any pythagorean triplet $(a,b,c)$, we know, by the definition that: $$a^2 + b^2 = c^2$$

But take $$a^x + b^x = c^x$$

Is $x=2$ the only possible solution $\in \Bbb R$ in this case? How can this be concluded?

I conjecture that $2$ is the only solution but I am not sure how to conclusively state this fact.

Answer 1

Divide by $c^x$, and get $$\left(\frac{a}{c}\right)^x+\left(\frac{b}{c}\right)^x=1$$ The left-hand side is a decreasing function of $x$. It equals 2 when $x=0$ and approaches 0 for large $x$, so there will be only one solution.
It is greater than $2$ when $x<0$ because $a$ and $b$ are less than $c$.

Answer 2

You should read up on Fermat's last theorem. It's not clear whether you want $x$ to be real or integer, but there is a section on exponents other than positive integers.