Prove that if $7^n-3^n$ is divisible by $n>1$, then $n$ must be even.

This is one of my favorite number theory problems because of the neat trick below :

Assume that $n$ is odd . Let $p$ be the smallest prime factor of $n$ (it's obviously odd and can't be $3$ or $7$)

From Fermat's little theorem :

$$7^{p-1}-3^{p-1} \equiv 1-1 \equiv 0 \pmod{p}$$

Also you know that :

$$p \mid n \mid 7^n-3^n$$

Now I'll use a standard lemma easily provable by induction :

Lemma If $a,b,n,m$ are positive integers and $(a,b)=1$ then : $$(a^n-b^n,a^m-b^m)=a^{(n,m)}-b^{(n,m)}$$

Use this lemma here so :

$$p \mid (7^{p-1}-3^{p-1},7^n-3^n)=7^{(n,p-1)}-3^{(n,p-1)}$$

Here comes the awesome part :

Because $p$ is the smallest prime factor of $n$ we must have $(p-1,n)=1$ . To see this assume that there is a prime $q$ such that $ q \mid n$ and $q\mid p-1$ this means that $q <p$ and $ q \mid n$ which contradicts the minimality of $p$

$(n,p-1)=1$ so we can simplify :

$$p \mid 4$$ which is a contradiction because $p$ is odd .

If $n\mid 7^n-3^n$ and $n>1$, then let $p$ be the least prime divisor of $n$.

Clearly $\gcd(21,p)=1$, so $\left(7\cdot 3^{-1}\right)^n\equiv 1\pmod{p}$, i.e. $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid n$.

By Fermat's Little theorem $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid p-1$. Therefore $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid \gcd(n,p-1)=1$, so $7\cdot 3^{-1}\equiv 1\pmod{p}$, so $7\equiv 3\pmod{p}$, so $p\mid 7-3=4$, so $p=2$.