Show that $70$ divide $101^{6n} - 1$ for all $n$ natural numbers.
I tried to show that $101^{6n}$$ \equiv 1$ mod $70$.
Thanks for all. I got it.
Note that $70$ $=$ $7$.$5$.$2$
As $101^{ϕ(7)}$ $\equiv 1$ mod $7$, so $101^{6n}$ $\equiv 1$ mod $7$ (Euler Theorem)
and $101^{ϕ(2)}$ $\equiv 1$ mod $2$, so $101^{6n}$ $\equiv 1$ mod $2$ (Euler Theorem)
and $101$ $\equiv 1$ mod $5$, so $101^{6n}$ $\equiv 1$ mod $5$
Therefore, $101^{6n}$ $\equiv 1$ mod $70$.
Since $101 \equiv 1\;(\text{mod}\;10)$, it follows that $101^{6} \equiv 1\;(\text{mod}\;10)$.
Since $101$ is not a multiple of $7$, it follows, by Fermat's Little Theorem, that $101^6\equiv 1\;(\text{mod}\;7)$.
Then $101^6 - 1$ is a multiple of both $10$ and $7$, hence it's a multiple of $70$.
Since $101^6\equiv 1\;(\text{mod}\;70)$, it follows that $101^{6n}\equiv 1\;(\text{mod}\;70)$, as was to be shown.