Use Fermat's Theorem to show that for every integer $n$, the number $n^{37}-n$ is divisible by $1919190$

Fermat's Little Theorem. If $a\in Z$ and $p$ is a prime not dividing $a$, then $p$ divides $a^{p-1}$, that is, $a^{p-1}\equiv1\pmod p$ for $a\not\equiv0\pmod p$.

I wish to show that the number $n^{37}-n$ is divisible by $1919190$. It means that $1919190$ divides $2^{37}-2, 3^{37}-3, 4^{37}-4$ etc.

Now applying Fermat's Little Theorem, I factored out $1919190$ into prime numbers: $(37)(19)(13)(7)(5)(3)(2)$.

So, I got stuck there and I don't know what to do next. Can you help me prove this problem or give me any alternatives.

Thanks for the help!

• Hint: consider $n^{37}-n\pmod p$ for all primes dividing $1919190$. – Wojowu Feb 21 '16 at 12:03
• Since no square of a prime number divides $1919190$, we have $1919190 \mid n^{37} - n$ if and only if $p \mid n^{37}-n$ for all primes $p$ dividing $1919190$. Now write $n^{37}-n = n\cdot (n^{36}-1)$, and apply Fermat's theorem for all primes dividing $1919190$. – Daniel Fischer Feb 21 '16 at 12:03
• @Wojowu I tried your hint but I'm not that confident with my proof. I'll gladly be thankful if you show me yours. – MJ. Rivo Feb 21 '16 at 15:38