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?
 A: Re write the Eq. as $$3^x-2^x=5^x-4^x....(1)$$
Use LMVT on $f(t)=t^x$ in $(2,3)$ and $(4,5)$
to write $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, t_1\in(2,3)~~~~(2)$$
and
$$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, t_2\in (4,5)~~~~(3)$$
so $t_1 \ne t_2$, By (1), we can equate $xt_1^{x-1}=xt_2^{x-1} \implies x=0~or~ x=1.$
So (1) has only two solutions: $x=0,1$.
A: 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
A: For $x\geq2$ we have $3^{x-2},4^{x-2}\leq5^{x-2}$, and so from $5^2=3^2+4^2$ it follows that
$$2^x+5^x>5^x=5^{x-2}(3^2+4^2)=5^{x-2}3^2+5^{x-2}4^2\geq3^x+4^x.$$
For $x\leq-2$ we have $3^{x+2},4^{x+2}\leq2^{x+2}$, and so from $2^{-2}>3^{-2}+4^{-2}$ it follows that
$$2^x+5^x>2^x=2^{x+2}(3^{-2}+4^{-2})=2^{x+2}3^{-2}+2^{x+2}4^{-2}\geq3^x+4^x.$$
Then it suffices to check that the equality holds for $x=0$ and $x=1$, but not for $x=-1$.
