A diophantine equation with unknown exponents: $ 4^n + 5^n = 7^m + 2^m $

Question

A friend proposed to me the following problem

Find all naturals $n$ and $m$ such that $ 4^n + 5^n = 7^m + 2^m $

I noticed that no solutions exist such that $m > n$ since then LHS < RHS.

When $m=n$, it seems that the only solutions are $(0,0)$ and $(1,1)$

The remaining case seems more challenging $ m < n$.

I haven't achieved any significant progress yet.

Would you please share any advice, suggestion, thanks in advance.

Answer 1

Claim: all solutions satisfy $m \le 1$.

Assume that there is a solution with $m \ge 2$. Looking at the equation modulo $4$ we get $$1 \equiv 7^m$$ thus, $m$ is even.
Now, looking at the equation modulo $5$, we get $$4^n \equiv 2\cdot 2^m$$ thus $2^{m+1} \equiv 4$ or $2^{m+1} \equiv 1$ mod $5$. but this is not possible for even $m$, since $2$ raised to an odd power can only be congruent to $2$ or $3$ mod $5$.

So the only solutions are $(0,0)$ and $(1,1)$.

Answer 2

Case $m=n$. By the Mean Value Theorem, there is $s\in (2,4)$ and there is $t\in (5,7)$ such that $$2ns^{n-1}=ns^{n-1}(4-2)=4^n-2^n=7^n-5^n=f'(t)=nt^{n-1}(7-5)=2nt^{n-1}$$ where $f(x)=x^n$. Now if $n>1$ then $2nt^{n-1}>2ns^{n-1}$ and it follows that there are no solutions for $n>1$.