Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$

$n\mid 7^n\!+6^n\!=:a_n$ odd $\Rightarrow n$ odd, so $\,a_n = 7^n\!-\!(-6)^n,\,$ so $\,b,c= 7,-6\,$ below $\,\Rightarrow\, p = 13$.

Lemma $ $ If $\,\gcd(b,c)\!=\!1\ $ & $\ 1<n\mid b^n\!-c^n$ then $\,p\mid b\!-\!c\,$ for $\,\color{#c00}{p}=$ least prime factor of $\,n$.

Proof $\,\ \color{blue}{p\nmid b}\,$ else $\,p\mid n\mid b^n\!-c^n\Rightarrow\,p\mid c,\,$ contra $\,\gcd(b,c)\!=\!1.\,$ $\color{#90f}{p\nmid c}\,$ by symmetry. If $\ \color{#0a0}{p\nmid b\!-\!c}\ $ $\rm\color{#90f}{then}$ $\!\bmod p\!:\ a:= \frac{b}c\!\not\equiv \color{#0a0}1,\color{blue}0,$ $\, a^n\equiv 1,\,$ so $\ {\rm ord}\ a\ge\color{#c00}p,\,$ contra $\ a^{p-1}\equiv 1\,$ by little Fermat, and the Order Theorem $\,(a^n\equiv1\,\Rightarrow\, {\rm ord}(a)\,$ is a factor of $\, n;\,$ but ${\rm ord}(a)\neq 1$ by $\color{#0a0}{a\neq 1}\,$ so the least value $\,{\rm ord}(a)\,$ can take is the least nontrivial factor of $\,n,\,$ which is its least prime factor $\,\color{#c00}p).$