Integer solutions for $2^x+5^x=3^x+4^x$ How many integer solutions exist for the equation $2^x+5^x=3^x+4^x$?
Attempt:
Clearly, $x=0$ is a solution.
Then, I divided both sides by $5^x$ to get;
$$\left(\frac 25\right)^x+1=\left(\frac 35\right)^x+\left(\frac 45\right)^x$$
It is easy to see that the left hand side of this equation is always greater than $1$. Moreover, taking a first derivative on the right hand side, it is easy to see that it is a decreasing function, which attains value $1$ at $x=2$. Thus, for $x>2$, RHS$<1$ while LHS$>1$, so no solutions exist, and the only natural solutions for $x$ are, trivially, $x=0,1$.
But I am struggling with the case where $x<0$. Can anyone please help with that? I also see that the bases on both sides have sum equal to $7$, which makes me think that some kind of inequality might be involved, but I am unable to proceed that way.
Edit: It would be helpful to see a solution that exploits the integer constraint.

Suppose there were an integer solution $x<0$. Then multiply both sides by $12^{-x}$ to get
$$30^{-x} + 12^{-x} = 20^{-x} + 15^{-x}$$
or
$$0 \equiv 1 \pmod 2,$$
a contradiction.
