Can you solve this?

Suppose that we can write $a^m+a-1=(a^n+a^2-1)Q(x)+R(x)$ , with $R(x)$ not identically zero. Then by the conditions, $R(x)$ must have infinitely many roots, contradiction; hence, $a^n+a^2-1|a^m+a-1$ . Since $0^n+0^2-1<0$ and $1^n+1^2-1>0, a^n+a^2-1=0$ has a root $0<r<1$ . Then also $r^m+r-1=0$ , so $r^m+r=r^n+r^2 \Rightarrow r(1-r)=r^n(1-r^{m-n}) \Rightarrow r(1-r)=(1-r^2)(1-r^{m-n})$ . Continuing, we have that $r=1-r^{m-n}+r-r^{m-n+1} \Rightarrow r^{m-n+1}+r^{m-n}=1$ . Since $r<1$ , we must have $m-n+1\le n$ or else $r^{m-n+1}+r^{m-n}< r^n+r^{n-1}\le r^n+r^2=1$ , contradiction. Hence, $m\le 2n-1$ .

Clearly, $a^n+a^2-1|(a^{m-n}(a^n+a^2-1)-(a^m+a-1)) \Rightarrow a^n+a^2-1|a^{m-n+2}-a^{m-n}-a+1$ . Since $a^{m-n+2}-a^{m-n}-a+1$ is not identically zero, and since $m\le 2n-1, m-n+2$ is either $m$ or $m+1$ . In the first case, we must have $a^{m-n+2}-a^{m-n}-a+1=a^n+a^2-1$ , but this is impossible. In the second case, we must have $(a-1)(a^n+a^2-1)= a^{n+1}-a^{n-1}-a+1 \Rightarrow -a^{n-1}=-a^n+a^3-a^2$ , which can only happen if $n=3$ , and therefore $m=5$ . As $(a^3+a^2-1)(a^2-a+1)=(a^5+a-1)$ , this solution works. So the only answer is $(m,n)=(5,3)$ .