, 2011

Write $p = 17^{2n} + 4 \equiv 4^n + 4 \pmod 5$ to deduce that $p \equiv 0 \pmod 5$ if $n$ is even. Since $p$ is prime, $n$ must be $0$, so $p = 5$ and the conclusion follows.

Henceforth assume $n$ odd. Rule out the case $n \equiv 5 \pmod 6$ on account of $p = 17^{2n} + 4 \equiv 3^n + 4 \pmod{13} \equiv 0 \pmod{13}$.

In the remaining cases, $n \equiv 1 \text{ or } 3 \pmod 6$, write $p = 17^{2n} + 4 \equiv 2^n + 4 \pmod 7$ to infer $p \equiv 5 \text{ or } 6 \pmod 7$, both of which are quadratic nonresidues modulo $7$; that is, $\left(\frac{p}{7}\right) = -1$.

Consequently, $\left(\frac{7}{p}\right) = -1$ by quadratic reciprocity, so $7^{(p-1)/2} \equiv \left(\frac{7}{p}\right) \equiv -1 \pmod p$. This ends the proof.