Prove that $n^7$ = n(mod 42)

Question

We have to prove $n^7$ = n(mod 42)

i tried as

We know $n^7$-n is divisible by 7

Hence n(n-1)(n+1)(n$^2$ +1 +n) is divisible by 7

Answer 1

Hint: $n^7-n=0$ mod 7 by little fermat

$n^7-n=n(n^3+1)(n^3-1)$ if $n$ is even, $n=0$ mod 3 done

$n$ is even $n=1$ mod 3, $n^3-1$ = 0 mod 3 done

$n$ is even, $n=-1$ mod 3, $n^3+1=0$ mod 3 done.

$n$ is odd, $n^3+1$ is even, if $n$=0 mod 3 done.

$n$ is odd, $n^3+1$ is even, $n=1$ mod 3, $n^3-1$ = 0 mod 3 done.

$n$ is odd, $n^3+1$ is even, $n=-1$ mod 3, $n^3+1$ = 0 mod 3 done.