Indices Mistake-2

Case 1: a=0

Obvious solutions are when $a=0$ , then the RHS is always $0$ . $\Rightarrow 0^{m-n} = 0$ . But we note that $0^{m-n}$ is undefined when $m-n \le 0$ . Then the acceptable cases are when $m > n$ . As $m, n \in [1,10]$ , then the number of acceptable cases is $m-1$ for an $m$ .

Then total number of cases for $a=0$ : $\displaystyle \sum _{m=1} ^{10} {(m-1)} = \sum _{i=1} ^{9} {i} = \dfrac {9(9+1)}{2} = 45$

Case 2: a = 2

We note that $a=1$ cannot have solution, because RHS is always $0$ but the LHS is always $1$ .

We note that for $a >1$ , $m > n$ is necessary for an integer solution. Therefore the minimum acceptable $m=2$ . When $a=2$ , LHS=RHS, when $m=2$ and $n=1 \Rightarrow 2^{2-1} = 2^2 - 2^1 = 2$ . For $m > 2$ , LHS < RHS. For $a > 2$ , LHS < for all $m \ge 2$ .

Therefore, there are altogether is $45 + 1 = \boxed{46}$ .