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.
 A: *

*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 $p\ne7$.
A: Since you've come up with the factorization, we can observe that this number is divisible by two other numbers, namely $n^3 + 4n^2 + 5n + 1$ and $n^2 + n +1$. The expression $(n+1)^5 + n$ prime if and only if one of the factors evaluates to one. But is this possible with positive $n$? (I'll leave this to you to explain why not.)
After that, a couple cases are possible.


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*Both factors of $P(n)$ are prime.

*One or more of the factors of $P(n)$ is composite. Then by the fundamental theorem of arithmetic, we can factor any composite number into a product of primes, which will yield two (possibly non-distinct) prime numbers that divide $P(n)$. We will prove below that there are at least two distinct prime numbers resulting from this.


There's only one last annoying thing to deal with. What if $P(n) = p^m$, for some prime $p$ and integer $m$? Then $P(n)$ would only be divisible by one unique prime $p$. But why can this not be the case? Well, if this were the case, since both $n^3 + 4n^2 + 5n + 1$ and $n^2 + n +1$ are integers, we would have to have $n^3 + 4n^2 + 5n + 1 = (n^2 + n +1)^q$, for some rational number $q$. We can easily verify this is not true. Taking into account the degree of these factors, we can conclude that the only possible candidate for $q$ is $q = 3/2$, because otherwise the degrees of both sides would not match up. From there, we can easily show that $n^3 + 4n^2 + 5n + 1 \neq (n^2 + n +1)^{3/2}$, so we are done.
A: $B$ and $C$ aren't necessarily prime, but they have distinct factors.  
To show this, continue as follows:
$(x^3+4x^2+5x+1)=(x^2+x+1)(x+3)+x-2.$
$x^2+x+1=(x-2)(x+3)+7.$
Therefore, $\gcd(x^3+4x^2+5x+1,x^2+x+1)|7,$ so there are two possibilities.
Case ($1$):  $\gcd(x^3+4x^2+5x+1,x^2+x+1)=1.$
In this case, since, for positive $x,$ $x^3+4x^2+5x+1>x^2+x+1>1$, 
$x^3+4x^2+5x+1$ and $x^2+x+1$ have distinct prime factors.
Case ($2$):  $\gcd(x^3+4x^2+5x+1,x^2+x+1)=7 .$
In this case $7|x-2.$  For $x=2, x^3+4x^2+5x+1=35$ has two distinct prime factors ($5$ and $7$).  
For $x>2$, $x^3+4x^2+5x+1>x^2+x+1>7$, and $\gcd\left(\dfrac {x^3+4x^2+5x+1}7,\dfrac{x^2+x+1}7\right)=1,$
so $\dfrac {x^3+4x^2+5x+1}7$  and $\dfrac{x^2+x+1}7$ have distinct prime factors.
