Solve in the set of integers $2^x +5^x = 3^x + 4^x$ 
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x +  4^x$$

With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$
So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.

If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).
Any easier solution?

You can show $5^n\ge 3^n+4^n$ for $n\ge2$ faster: $5^2\ge 3^2+4^2$ and in the induction step
$$ 5^{n+1}=5\cdot 5^n\ge 5\cdot3^n+5\cdot4^n>3^{n+1}+4^{n+1}.$$
An alternative idea for negative $x$ s also by size comparison with $2^{-n}$ being quite big
