Peter wrote: Show that there are infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is divisible by $n$. We prove that for any positive integer $m$ the number $n=3^{2^{m}}-2^{2^{m}}$ satisfies the problems condition. Indeed, we have that if $a|b$ then $3^{a}-2^{a}|3^{b}-2^{b}$ so to prove that $n|3^{n-1}-2^{n-1}$ it is sufficent to prove that $2^{m}|n-1$,that is $2^{m}|(3^{2^{m}}-1)-2^{2^{m}}$. Since $2^{m}\geq m$, then $2^{m}|2^{2^{m}}$ and we only need to verify that $2^{m}|3^{2^{m}}-1$. It is easy to see that \[3^{2^{m}}-1=(3-1)(3+1)(3^{2}+1)...(3^{2^{m-1}}+1)\] which shows that $3^{2^{m}}-1$ is a product of $8=(3-1)(3+1)$ and $m-1$ even numbers and thus is divisible by $2^{m}$(actually we have proved that it is divisible by $2^{m+2}$).

I will post bellow a solution for a generalization of this problem. Let $a$ and $b$ be positive integers such that $a>b$ and they are coprime. Show that there are infinitely many composite numbers $n$ such that $a^{n-1}-b^{n-1}$ is divisible by $n$. Solution: We take an arbitrary prime number $p$ such that $a^2-b^2$ isn't divisible by $p$. We will show that the number $n = \frac{a^{2p}-b^{2p}}{a^2-b^2}$ satisfies the problem's condition. $n = \frac{a^{2p}-b^{2p}}{a^2-b^2} = \frac{a^{p}-b^{p}}{a-b}\cdot\frac{a^{p}+b^{p}}{a+b}\newline = (a^{p-1}+a^{p-2}b+\cdots+b^{p-1})(a^{p-1}-a^{p-2}b+\cdots+b^{p-1})$. Obviously,we have $a^{p-1}+a^{p-2}b+\cdots+b^{p-1}>a^{p-1}-a^{p-2}b+\cdots+b^{p-1}>1$. Therefore $n$ is a composite number. $(a^2-b^2)n \equiv a^{2p}-b^{2p} \equiv a^2-b^2 \pmod{p}$. Since $a^2-b^2$ and $p$ are coprime, we have $n \equiv 1 \pmod{p}$. I claim that $n$ is an odd number. Otherwise, $n$ is an even number. The equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p}-b^{2p}$ is an even number. So we have that $a$ and $b$ have same parity. Since they are coprime, they are odd numbers. But we have $n \equiv a^{2p-2}+a^{2p-4}b^2+\cdots+b^{2p} \equiv 1+1+\cdots+1 \equiv p \equiv 1 \pmod{2}$. This is a contradiction. From the above, $n-1$ is divisible by $2p$. Or there is a positive integer $k$ such that $n-1=2pk$. The equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p} \equiv b^{2p} \pmod{n}$. Therefore, we have $a^{n-1} \equiv a^{2pk} \equiv b^{2pk} \equiv b^{n-1} \pmod{n}$. So,we have done.