North Macedonia

Let $13|7^n+n^5$. Clearly $13$ does not divide $n$, so by Little Fermat's theorem we have that $n^{12}\equiv 1(\mathrm{mod}13)$, i.e. $13|n^{12}-1$.

Now we have $13|7^n+n^5$, i.e. $13|n^7(7^n+n^5)$ or $13|n^7{7}^n+1+n^{12}-1$, and because of $13|n^{12}-1$ we get $13|n^7\cdot 7^n+1$.

Now let $13|n^7\cdot 7^n+1$; clearly $13$ does not divide $n$, so as previously we have $13|n^{12}-1$. We get $13|n^7\cdot 7^n+1$, i.e. $13|n^5(n^7\cdot 7^n+1)$, i.e. $13|n^{12}\cdot 7^n+n^5$, or $13|(n^{12}-1)7^n+7^n+n^5$, from where we have that $13|7^n+n^5$, which concludes the proof.