<s>[INST] show that $7\mid p^3-p$ if $p$ is a prime divisor of $n^3+n^2-2n-1$ 
Let $p$ be a prime number, and $n$ a positive integer such
$$p\mid n^3+n^2-2n-1, \quad n\ge 2.$$
Show that $$7\mid p^3-p.$$

It maybe can use  Fermat's little theorem?
 [/INST] An alternative way is to consider the prime divisors of
$a^3+a^2b-2ab^2-b^3$, and reduce the numbers by applying Thue's lemma.
Call a prime $p$ "bad" if $p\not\equiv0,\pm1\pmod7$, and there exist some co-prime integers $a,b$ such that $p$ divides $a^3+a^2b-2ab^2-b^3$. Suppose that there at least one bad prime, and take the least bad prime and the corresponding $a$ and $b$.
By Thue's lemma, there are some integers $0<k<p,c,d$ such that $ak\equiv c\pmod{p}$, $bk\equiv d\pmod{p}$, $|c|<\sqrt{p}$, and $|d|<\sqrt{p}$. Then $p$ divides $c^3+c^2d-2cd^2-d^3$. If $c$ and $d$ are not co-prime, we can divide them by their gcd. (Notice that at least one of $c$ and $d$ is not divisible by $p$, so this simplification does not destroy the divisibility by $p$.)
It can be verified that $c^3+c^2d-2cd^2-d^3\equiv\pm1\pmod7$ or $c^3+c^2d-2cd^2-d^3\equiv\pm7\pmod{49}$. Any case happens, there must be another "bad" prime divisor $q$ of $c^3+c^2d-2cd^2-d^3$. But $p$ is the least bad prime, so $q\ge p$; therefore $p^2\le pq\le|c^3+c^2d-2cd^2-d^3|<3\sqrt{p}^3$, so $p<9$. 
The primes $2,3,5$ can be excluded manually.
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