On the prime numbers Let $p=q^2+q+1$ where p,q are prime numbers. Is it true that $p|(q−1)^2 (q^2+q)$?
MY TRY: 
I said that since $(p=q^2+q+1 , q^2+q)=1$, we have $p=q^2+q+1 \mid (q-1)^2$. Since $q^2+q+1  > (q-1)^2 $ , we get a contradiction, that is $p \nmid (q-1)^2 (q^2+q)$.
Now i want to know is it true?
 A: $q^2+q\equiv {-1} \pmod p $  and $p=(q-1)^2+3q\implies (q-1)^2\equiv {-3q}\pmod p$ 
Combining these two we get $p|3q$ and thus $p|q$ or $p|3$. Now former one is impossible since $p>q$ and latter being impossible as $p=3\implies q=1$.
A: HINT.- You can reduce your problem to just one prime because you have the question
$$\frac{q^3-1}{q-1}\space |\space(q-1)^2(q^2+1)$$ where $q$ is prime.
This way you can maybe be sure yourself of your good reasoning.
A: Actually $p=q^2+q+1$ does not divide $(q-1)^2(q^2+q)$ even without the primality assumptions.  All we need to assume is that $q\gt1$.
If $(q-1)^2(q^2+q)=(q^2+q+1)k$, then $0\equiv k$ mod $q$, so $k=qh$ with $h\ge0$. This gives $(q-1)^2(q+1)=(q^2+q+1)h$, so $h\equiv1$ mod $q$, or $h=qr+1$ with $r\ge0$. Plugging this in leads to
$$(r-1)q^2+(r+2)q+(r+2)=0$$
which implies $r+2=qn$ with $n\gt0$.  Plugging this in leads to
$$nq^2+(n-3)q+n=0$$
The left hand side is clearly positive if $n\ge3$. For $n=2$, the quadratic $2q^2-q+2=0$ has no real roots.  And $q^2-2q+1=0$ has $q=1$ as its only solution.
There might be some easier way to organize the proof.  If so, I'd like to see it.
Added later:  Oh damn, Vidyanshu Mishra's answer simplifies things enormously.  The conclusion there that $p\mid3q$ does not depend on $p$ or $q$ being prime, and it immediately gives $3q=pn=(q^2+q+1)n$, or $nq^2+(n-3)q+n=0$.
