North Macedonia competition problems

Let $13|7^n + n^5$. Clearly $13$ does not divide $n$, so by Little Fermat's theorem we have that $n^{12} \equiv 1 \pmod{13}$, 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.