Where's the increasing function?

Let $n = x - 2$ . Then $x = n + 2$ and when $x > 2$ , we have $n > 0$ .

Then $3^x + 4^x = 3^{n + 2} + 4^{n + 2} = 3^n3^2 + 4^n4^2$

and $5^x = 5^{n + 2} = 5^n5^2 = 5^n(3^2 + 4^2) = 5^n3^2 + 5^n4^2$ .

For all $n > 0$ , $3^n < 4^n < 5^n$ , so $3^n3^2 < 5^n3^2$ and $4^n4^2 < 5^n4^2$ , and so $3^n3^2 + 4^n4^2 < 5^n3^2 + 5^n4^2$ which means $3^x + 4^x < 5^x$ for all $x > 2$ .

Therefore, there are no solutions $x > 2$ (integer or non-integer) for the equation $3^x + 4^x = 5^x$ .

Let $f(x) = \left ( \dfrac{3}{5} \right )^x + \left ( \dfrac{4}{5} \right )^x.$ Differentiating, we have

$\begin{aligned} f'(x) &= \left ( \dfrac{3}{5} \right )^x \ln \dfrac{3}{5} + \left ( \dfrac{4}{5} \right )^x \ln \dfrac{4}{5} \\ &= - \left [ \left ( \dfrac{3}{5} \right )^x \ln \dfrac{5}{3} + \left ( \dfrac{4}{5} \right )^x \ln \dfrac{5}{4} \right ] \end{aligned}$

The expression in the brackets will always be positive, since the exponential function is always positive and $\ln \dfrac{5}{3}, \ln \dfrac{5}{4} > \ln 1 = 0.$ Therefore, for all real values of $x,$ we have $f'(x) < 0,$ so $f$ is a monotonically decreasing function. This means that there can only be one value of $x$ such that $f(x) = 1,$ or in other words, there's only one real solution to the equation $3^x + 4^x = 5^x.$ Since we know that $x = 2$ is a solution, it is the only solution, so we conclude that no , there are no more solutions to the equation greater than 2.