To the 32nd power!

First, we shall prove that, for any relatively prime positive integers $x, y$ and any positive integer $n,$ any odd prime divisor of $x^{2^n} + y^{2^n}$ must have the form $2^{n + 1}k + 1$ for some positive integer $k.$

Let $p$ be an odd prime divisor of $x^{2^n} + y^{2^n}.$ We have

$\begin{aligned} x^{2^n} + y^{2^n} &\equiv 0 \! \! \! \! \pmod{p} \\ x^{2^n} &\equiv -y^{2^n} \! \! \! \! \pmod{p} \\ (xy^{-1})^{2^n} &\equiv -1 \! \! \! \! \pmod{p} \\ (xy^{-1})^{2^{n + 1}} &\equiv 1 \! \! \! \! \pmod{p}. \end{aligned}$

Note that $y^{-1}$ exists because $p$ is a prime and $y$ cannot be divisible by $p.$ If $y$ were divisible by $p,$ then $x$ would also have to be as well, which violates our assumption that $x, y$ were relatively prime.

Let $d = \text{ord}_p(xy^{-1}).$ We have $d|2^{n + 1}$ by what we just derived, and $d|p - 1$ by FLT. We also have $d \nmid 2^n,$ since otherwise we would have $(xy^{-1})^{2^n} \equiv 1 \! \pmod{p},$ forcing $p = 2,$ which is not valid. Therefore, $d = 2^{n + 1}.$ Since $d|p - 1,$ this means $p = 2^{n + 1}k + 1,$ and we are done. $\blacksquare$

Now, let's apply this result to the problem. We have $x = a, y = 1, n = 5.$ Thus, any odd prime divisor of $a^{32} + 1$ is of the form $64k + 1.$ However, $64 \nmid 2016 = 2017 - 1,$ so 2017 cannot be written in this form. Thus, no values of $a$ satisfy the problem statement.

Suppose that $a^{32}+1$ is divisible by 2017, i.e. $a^{32}\equiv -1\pmod{2017}$ . We could restrict the possible value of $a$ in the range $[1,2016]$ and because $2017$ is prime we also have $\gcd(a,2017)=1$ . Note that $\phi(2017)=2016=63*32$ and, due to Euler's Theorem $a^{2016}\equiv 1\pmod{2017}$ . But from $a^{32}\equiv -1\pmod{2017}$ we also have $1= a^{2016}=(a^{32})^{63}\equiv (-1)^{63}=-1\pmod{2017}$ and so $1\equiv -1\pmod{2017}$ which is impossible.