<s>[INST] A question on Fermat's Little Theorem Let $n$ be an integer not divisible by $3$. Show that $n^7 ≡ n (\mod 63)$.
I know that we can split $63$ into $3^2 \cdot 7$ 
So we have $n^7=n (\mod 7\cdot3^2)$
$n^7=n (\mod 3^2)$ and $n^7 = n (\mod 7)$
And I am stuck how to go about solving this question after this
 [/INST] $$n^7-n=n(n^3-1)(n^3+1)=(n-1)n(n+1)(n^2-n+1)(n^2+n+1).$$
Now, it's obvious that $(n-1)n(n+1)$ is divisible by $3$.
Let $n=3k-1$, where $k$ be a natural number.
Thus, $n^2-n+1$ is divisible by $3$.
Let $n=3k+1$, where $k$ be a natural number.
Thus, $n^2+n+1$ is divisible by $3$,
which says that $n^7-n$ is divisible by $9$.
Also, $n^7-n$ is divisible by $7$ for all integer $n$ by the Fermat's little theorem, which says that it's divisible by $7$ for all integers $n$, which not divisible by $3$ and we are done! 
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