I found this problem on some training material for a sixth grade Romanian math competition, and I literally have no clue how to approach it. I am not even sure if my interpretation of the problem text is correct, so if someone has a counter example I'll be glad to see it.
Proof that there is no $a, b, n \in N, b \ge 2 $ such $a^{b} = 2017^{n} +43$. My only try was to look at the last digit of $2017^{n} +43$" and all I could figure it out was that for $n=4k +2$ last digit will be two, so in this case the number can't be a perfect square
Thank you
If $a,b,n\in\Bbb{N}$ are such that $a^b=2017^n+43$ then in particular \begin{eqnarray*} a^b&\equiv&2017^n+43\equiv4\pmod{8},\\ a^b&\equiv&2017^n+43\equiv2\pmod{3}, \end{eqnarray*} where the first implies $b=2$ and the second implies $b\neq2$, a contradiction.