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?

A blind approach, although it will give the same solution.
The polynomial $f(x)=x^3+ x^2 - 2 x -1$ has discriminant $\Delta=49$, a square. Therefore, given a root $x_1$ of $f(x)$, the other roots can be expressed in terms of $x_1$ from the equality
$$(x_2-x_3) = \frac{(x_1-x_2)(x_1-x_3)(x_2-x_3)}{(x_1-x_3)(x_2-x_3)}= \frac{\pm\sqrt{\Delta}}{f'(x_1)}$$
and combined with the equality $x_2+x_3= s-x_1= -1-x_1$ give the expression for the other roots. 
Note that we have a GCD formula for $f$ and $f'$ involving the discriminant. In our case we have 
$$(-6 x -1)(x^3+x^2- 2 x -1)+ (2x^2 + x- 3)(3 x^2 + 2 x - 2)=7$$
and so if $\alpha $ is a root of $f$ then 
$$\frac{\sqrt{\Delta}}{f'(\alpha)}= \frac{7}{f'(\alpha)} = 2\alpha^2 + \alpha- 3$$
Therefore, if $\alpha$ is a root of $f$ in a field $F$ then 
$$\frac{1}{2}(-1-\alpha \pm (2\alpha^2 + \alpha- 3))= \alpha^2 - 2, - \alpha^2 - \alpha + 1$$
are also roots of $f$. Also,the product of their differences is $7$, so they are distinct if $7 \ne \operatorname{char} (F)$. In that case, starting from $\alpha$ a root we get all the roots in this way. 
This works for any cubic with discriminant a square. This one is special, because we notice:
$$\alpha \text{ root of }\ f \implies \alpha^2-2 \text{ root of }\ f$$
and we might recall the polynomial $x^2-2$ from $t^2 +\frac{1}{t^2}=(t+\frac{1}{t})^2-2$. Now we check that 
$$- \alpha^2 - \alpha -1 = (\alpha^2-2)^2 -2$$ if $\alpha$ is a root. 
So we may rephrase the above statement as (assume $7 \ne \operatorname{char}(F)$
$$t+\frac{1}{t} \text{root of }\ f \implies t^2 +\frac{1}{t^2} , t^4 +\frac{1}{t^4}\text{the other roots of }\ f$$
Now consider $p\ne 7$ such that $f$ has a root $\alpha$ in $F_{p}$. Consider then $t$ in $F_{p^2}$ such that $t+\frac{1}{t}=\alpha$. Then 
$$t+\frac{1}{t}, t^2+\frac{1}{t^2}, t^4+\frac{1}{t^4}$$ are the distinct roots of $f$ , all in $F_{p}$.  But then $t^8 + \frac{1}{t^8}$ must get us back to  $t + \frac{1}{t}$. So... $t^8=t$ or $t^8=\frac{1}{t}$. First case gives us an element of order  $7$ in $F^{\times}_{p^2}$ , so $7\mid p^2-1$.  Second case we hope to exclude. Short of other ideas, we substitute:
$$f(t+\frac{1}{t})=\frac{t^6 + t^5 + t^4 + t^3 + t^2 + 1}{t^3}$$
Therefore 
$$t^6 + t^5 + t^4 + t^3 + t^2 + 1 = 0$$ and 
we get $t^7=1$, $t\ne 1$, so $t$ element of order $7$ in $\mathbb{F}_{p^2}^{\times}$.
Added: another approach:  we have 
$$1+ (t+ \frac{1}{t}) + (t^2+ \frac{1}{t^2}) + (t^4 + \frac{1}{t^4}) = 0$$
so 
$$\frac{(t+ \frac{1}{t}-1)(t^6 + t^5 + t^4 + t^3 + t^2 + 1)}{t^3}=0$$
