How to show that $3^x+4^x=5^x$ has only one solution?

Question

How to show that $3^x+4^x=5^x$ has only one solution? Thanks in advice.

Answer 1

Define $$f(x)=\left(\dfrac{3}{5}\right)^x+\left(\dfrac{4}{5}\right)^x.$$ It is clearly monotonic decreasing. Note also that $$f(2)=1,$$ so that $$x=2$$ is the only real solution to $f(x)=1$.

More generally, for the equation

$$3^x+4^x+5^x=6^x,$$ define $$f(x)=\left(\dfrac{3}{6}\right)^x+\left(\dfrac{4}{6}\right)^x+\left(\dfrac{5}{6}\right)^x.$$ This function is also decreasing, and as Euler first noted, $$3^3+4^3+5^3=6^3,$$ so that $$f(3)=1,$$ and $x=3$ is the only real solution to $f(x)=1$.

Answer 2

Indeed $x = 2$ is a solution to $3^x + 4^x = 5^x$. We can verify that $x = 0$ and $x = 1$ don't work. So any other solutions must be greater than two. But, we have

$$3^x + 4^x = 5^x, \quad x > 2$$ which has no solutions by Fermat's Last Theorem.

(This is assuming that you are looking for integral solutions, as I asked you in the comments)