<s>[INST] 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?
 [/INST] 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: $x=0,1$.
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