<s>[INST] Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$ Prove: if $n\mid 7^n+6^n$ and $n>1$, then $13\mid n$
Let $p$ be the least prime number such that $p\mid n$. 
And I want to show that $p=13$
Let $d$ be the least number such that: $14^d\equiv 0 \pmod {p}$
And by Fermat's little theorem I have: $14^{p-1}\equiv 0 \pmod{p}$
Here I'm stuck 
 [/INST] OP poses two questions: If $n\mid 7^n+6^n$, first show $13\mid n$, and second show that $13$ is the smallest prime factor of $n$
Start by noting that any prime factor of $n$ cannot be a prime factor of either $7^n$ or $6^n$, as a prime factor which divides $n$ and either of $7^n$ or $6^n$ must also divide the other. That is impossible because $\gcd(7^n,6^n)=1$. Thus, none of $2,3,7$ can be factors of $n$. In particular, this means that $n$ must be odd, a fact which is also apparent because the sum $7^n+6^n$ is odd.
Next, recall that for any odd exponent $k$, it is the case that $(a+b)\mid (a^k+b^k)$. Since $n$ is restricted to being odd, this means that $(7+6)\mid (7^n+6^n)$. This shows that $13\mid n$, answering OP's first query.
Since we know that $2,3,7$ are not divisors of $n$, we need only show that $5$ and $11$ are not divisors of $n$ to establish that $13$ is the smallest prime divisor of $n$
The case of $5$ is simple: $7^n+6^n \equiv 2^n+1^n \bmod 5$, and $2^n+1^n \equiv 0 \bmod 5 \iff n=4k+2$. That requires $n$ to be even, and $n$ is restricted to being odd, so $5\mid n$ is ruled out.
The case of $11$ takes a bit more work. Working modulo $11$, and recalling that when a modulus $p$ is prime, $a^k \bmod p \equiv a^{k \bmod (p-1)} \bmod p$
$$\begin {array}{c|c|c|c|c|c|c|c|c|c|c|c|} \\
n\equiv &1&2&3&4&5&6&7&8&9&10& \bmod 10 \\
6^n \equiv &6&3&7&9&10&5&8&4&2&1& \bmod 11 \\
7^n \equiv &7&5&2&3&10&4&6&9&8&1& \bmod 11 \\
\sum \equiv &2&8&9&1&9&9&3&2&10&2& \bmod 11 \\
\end {array}$$
Since the sum $7^n+6^n \not \equiv 0 \bmod 11$ for all $n$, it is the case that $11 \not\mid n$.
This shows that no prime smaller than $13$ divides $n$, answering OP's second query.
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