Determine all pairs $(a, b)$ of non-negative integers that satisfy

$$
2017^{a}=b^{6}-32 b+1
$$

(Walther Janous)
Answer: The two solution pairs are $(0,0)$ and $(0,2)$.

Since $2017^{a}$ is odd, $b$ must be even, so $b=2c$, where $c$ is an integer. Consequently, $2017^{a}=64(c^{6}-c)+1$, so $2017^{a} \equiv 1 \pmod{64}$. However, $2017 \equiv 33 \pmod{64}$ and $2017^{2} \equiv (1+32)^{2} = 1 + 2 \cdot 32 + 32^{2} \equiv 1 \pmod{64}$, so the powers of 2017 modulo 64 alternate between 1 and 33. Therefore, $a$ must be even. Thus, $2017^{a}$ is a perfect square. We denote the polynomial on the right-hand side of the equation by $r(b) = b^{6} - 32b + 1$ and show that for $b > 4$, it lies between two consecutive perfect squares.

Let $b > 4$. We have $r(b) > 0$. Additionally, $r(b) > (b^{3} - 1)^{2}$, because $b^{6} - 32b + 1 > b^{6} - 2b^{3} + 1 \Leftrightarrow b > 4$. Since the perfect square $2017^{a}$ is supposed to lie between two consecutive perfect squares, there are no solutions in this case.

Since $b$ is even, we only need to check $b=4$, $b=2$, and $b=0$.

For $b=4$, we see immediately that the equation modulo 3 becomes $1 \equiv 1 - 2 + 1 = 0$, so this does not yield a solution.

For $b=2$, we have $2017^{a} = 2^{6} - 2^{6} + 1$, so we obtain the solution pair $(a, b) = (0, 2)$.

For $b=0$, we have $2017^{a} = 1$, so we obtain the solution pair $(a, b) = (0, 0)$.

These are the only solutions.

(Walther Janous)