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.
 A: I think the solutions in the link provided answer a different question--nonintegral $x$, and so are not helpful here. For $x$ integral, there is a much simpler observation.
For $x$ a negative integer, note that $2^x > 3^x+4^x$ for all integers $x \le -3$ [you can show this by induction], so if $x$ is nonpositive, then $|x|$ must be no larger than $2$.
Likewise for $x$ a positive integer, note that $5^x > 3^x+4^x$  for all $x \ge 3$ [you can show this by induction as well], so $x$ must also be no larger than $2$.
So all this leaves you to check is which $x \in \{-2,-1,0,1,2\}$ satisfy this equation.
A: 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.
A: If $x < 0$ replace $x$ with $-y$ and get
$2^{-y} + 5^{-y} = 3^{-y} + 4^{-y}$ and
$(\frac 12)^y + (\frac 15)^y = (\frac 13)^y + (\frac 14)^y$ and to get fractions $\le 1$ multiply all sides by $2^y$ and do what you did before.
$1 + (\frac 25)^y \ge 1$ and $(\frac 23)^y + (\frac 12)^y$ is, by derivative, a decreasing function.
For $y = 1$ we don't have a solution ($1+\frac 25 \ne \frac 23 + \frac 12$) and for $y \ge 2$ we have $(\frac 23)^y + (\frac 12)^y \le \frac 49 +\frac 14 < 1 < 1+(\frac 25)^y$.
So no negative integer solutions either.
So only integer solution is $x = 0$.
