Prove that there are infinitely many postive integers $n$ such that $2^n-8$ is divisible by $n$, and $n$ has least three distinct prime factors.
I only find infinitely many postive integers $n$ such that $2^n-8$ is divisible by $n$,
such $n=3p$, use $$2^{3p}-8=8(2^{3(p-1)}-1)$$ Use Fermat theorem we have $$2^{p-1}-1\equiv 0\pmod p$$ and it is clear $$2^{3p}-8\equiv (-1)^{3p}-2\equiv 0\pmod 3$$ But this example is not such $n$ has least three distinct prime factors.so How find it? Thank you
PS:This problem is from Croatia Mathematical Olympiad exam (2013 or 2014)