Brazilian Math Olympiad

Let $p = 60k + 7$ be such a prime. Then $10^{2n} + 8 \cdot 10^n + 1 \equiv 0 \pmod{p} \iff (10^n - 1)^2 \equiv -10^{n+1} \pmod{p}$. Now suppose $n$ is odd. Then $(10^n - 1)^2 \equiv -1 \cdot (10^{(n+1)/2})^2 \pmod{p}$, and $-1$ is a quadratic residue. But by the Euler criterion, $\left(\frac{-1}{p}\right) = (-1)^{(p-1)/2} = -1$, a contradiction. So $n$ is even and, moreover, $\left(\frac{-10}{p}\right) = 1 \iff \left(\frac{2}{p}\right)\left(\frac{5}{p}\right) = -1 \iff \left(\frac{2}{p}\right) = -\left(\frac{5}{p}\right)$.
By the quadratic reciprocity lemma, $\left(\frac{5}{p}\right)\left(\frac{p}{5}\right) = (-1)^{\frac{5-1}{2} \cdot \frac{p-1}{2}} = 1 \iff \left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = \left(\frac{2}{5}\right) = -1$. So $\left(\frac{2}{p}\right) = 1 \iff (-1)^{\frac{p^2-1}{8}} = 1$, so $\frac{p^2-1}{8} = (30k+3)(15k+2)$ is even. Since $30k+3$ is odd, $15k+2$ is even, that is, $k$ is even.