shortlistBMO 2011

Write $p=17^{2n}+4\equiv 4^n+4(\mathrm{mod}5)$ to deduce that $p\equiv 0(\mathrm{mod}5)$ 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 $n\equiv 5(\mathrm{mod}6)$ on account of $p=17^{2n}+4\equiv 3^n+4(\mathrm{mod}13)\equiv 0(\mathrm{mod}13)$.

In the remaining cases, $n\equiv 1\text{ or }3(\mathrm{mod}6)$, write $p=17^{2n}+4\equiv 2^n+4(\mathrm{mod}7)$ to infer $p\equiv 5\text{ or }6(\mathrm{mod}7)$, 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 $7^{(p-1)/2}\equiv \left(\frac{7}{p}\right)\equiv -1(\mathrm{mod}p)$. This ends the proof.