Saudi Arabian Mathematical Competitions

We will prove that if $a>1$ is a positive integer such that $a-1$ is not a power of $2$, then there are infinitely many positive integers $n$ such that $n$ divides $a^{a^{n}-1}-1$ but $n$ does not divide $a^{n}-1$. Therefore, the given problem follows directly as an application with $a=2017$.

In fact, there is an odd prime divisor $p$ of $a-1$. Put $s=v_{p}(a-1) \geq 1$, for any positive integer $k$, by LTE we have $v_{p}\left(a^{p^{k}}-1\right)=v_{p}(a-1)+k$, this means that $n=\frac{a^{p^{k}}-1}{p^{s+1}}$ is a positive integer. By LTE again,
$$v_{p}\left(a^{n}-1\right)=v_{p}(a-1)+v_{p}(n)=s+v_{p}(a-1)+k-(s+1)=s+k-1 \geq k.$$
Thus, $p^{k} \mid a^{n}-1$, we get $n\mid a^{p^{k}}-1 \mid a^{a^{n}-1}-1$. It remains to show that $n \nmid a^{n}-1$. By above, we can write $a^{n}-1=p^{s+k-1} A$ and $n=p^{k-1} B$ for some positive integers $A, B$ coprime to $p$. This gives
$$\left(n, a^{n}-1\right)=\left(p^{k-1} A, p^{k-1} B\right)=\frac{1}{p^{s}}\left(p^{s+k-1} A, p^{s+k-1} B\right)=\frac{1}{p^{s}}\left(\frac{a^{p^{k}}-1}{p}, a^{n}-1\right)$$
which is clearly a divisor of
$$\frac{1}{p^{s}}\left(a^{p^{k}}-1, a^{n}-1\right)=\frac{1}{p^{s}}\left(a^{\left(p^{k}, n\right)}-1\right)=\frac{1}{p^{s}}\left(a^{p^{k-1}}-1\right)<\frac{a^{p^{k}}-1}{p^{s+1}}=n.$$
This shows that $n$ cannot divide $a^{n}-1$. The proof is done. $\square$