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 $s=v_p(a-1)\ge 1$, 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, $$v_p\left(a^n-1\right)=v_p(a-1)+v_p(n)=s+v_p(a-1)+k-(s+1)=s+k-1\ge k.$$ 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. $\square$