Prove that for odd $n > 1$ , $3^{n} + 1$ is not divisible by $n$.
There's a hint but I can't find any use for that.
hint: If $a$ and $b$ are coprime with $m$ and $a^{x} \equiv b^{x}$ (mod $m$) and $a^{y} \equiv b^{y}$ (mod $m$) then $a^{\gcd(x,y)} \equiv b^{\gcd(x,y)}$ (mod $m$).
I don't know if you can find any use for this but this is my thoughts : $3$ and $n-1$ are coprime with $n$ and $3^{n} \equiv (n-1)^{n} \equiv -1$ (mod $n$).
Let $n \mid (3^n+1)$ be odd. Then, you have $n \mid (3^{2n}-1)$. Let $p$ be the smallest prime factor of $n$. We know that $p \mid (3^{p-1}-1)$ by Fermat's Little Theorem (clearly, $p \neq 3$). This shows: $$p \mid (3^{\gcd(2n,p-1)}-1) $$
However, as $p$ is the smallest prime factor of $n$, $2n$ cannot be sharing any odd prime factors with $p-1$. Moreover, $2$ is a common factor of $2n$ and $p-1$ but as $n$ is odd, we have $4 \nmid 2n$. This shows: $$\gcd(2n,p-1)=2 \implies p \mid(3^2-1) \implies p\mid 8$$
But how can $p$ be an odd prime? Contradiction!
