Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$


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*Find the quotient of $P(x)$ divided by $x^2 + x + 1$

*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers. 
I've done the first task and came up with $P(x)= (x+1)^5 + x = (x^3+4x^2+5x+1)(x^2+x+1)$ which can be written as $A =BC +D$ where $D=0$. So, for the second task I think I have to prove that B or C or both are prime, but I don't know how to continue.

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*If $a,b$ are a roots of $x^2+x+1=0$ then $a^3=b^3 =1$ and by Vieta we have $a+b=-1$, so $a+1 =-b$. Now since $$p(a) = (a+1)^5+a = -b^5+a = -b^2+a = (b+1)+a = 0$$
and the same for $p(b)=0$ we see that $x^2+x+1$ divide $p$, so the remainder is $0$.





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*Since for $n\geq 2$ we have $p(n)=\underbrace{(n^2+n+1)}_{\geq 7}\underbrace{(n^3+4n^2+5n+1)}_{\geq 35}$
both parts are divisible by some prime. 
Let $d$ be a gcd of both factors. So $d\mid n^2+n+1$ and $d\mid n^3+4n^2+5n+1$ then $d\mid n^3-1$ and so $d\mid n-2$ so $d\mid 7$. So at least one factor must be also divisible by some prime $p\ne7$.
