Question

If $n$ is any prime number greater than $3$, except $7$, show that $n^6-1$ is divisible by $168$.

Answer 1

$n^6-1=(n^3+1)(n^3-1)=(n+1)(n-1)(1+n+n^2)(1-n+n^2)$

As $n$ is a prime number greater than $2$, $(n+1)$ or $(n-1)$ has to be divisible by $3$.

Because out of $3$ consecutive numbers one would be divisible by $3$ and certainly $p$ can't be that number.

$n+1$ and $n-1$ both have to be even because $n$ being prime greater than $3$ can't be even.

And out of two consecutive even numbers one has to be divisible by $4$.

Now, either $(1+n^2+n^4)$ or $(1-n^2+n^4)$ has to be divisible by $7$ for $n\ne 7$.

Hence, $n^6-1$ would be divisible by $3\times 2\times 4\times 7=168$.