Saudi Arabian Mathematical Competitions, 2015

We will look for integers of the form $n = 7^{a} - 3^{a}$ with $a$ dividing $n-1$. Clearly, if $a$ exists then $n = 7^{a} - 3^{a}$ divides $7^{n-1} - 3^{n-1}$.

Let $a = 3^{r}$ for $r \geq 1$. We have $n = 7^{a} - 3^{a} \equiv (-1)^{3^{r}} - 3^{3^{r}} \equiv 4 \pmod{8}$. We deduce that $n$ is a composite number.

On the other hand, because $3$ divides $7-1$, we have from the Lifting The Exponent theorem that $v_{3}\left(7^{a} - 1\right) = v_{3}(7-1) + v_{3}(a) = r+1$. We deduce that $a = 3^{r}$ divides $7^{3^{r}} - 1$ and therefore it divides $n-1 = 7^{3^{r}} - 1 + 3^{3^{r}}$. But there are infinitely many such $a$. This solves the problem.