Prove or disprove $n^2+41n+41$ is a prime number for every integer $n$ 
Prove or disprove $n^2+41n+41$ is a prime number for every integer $n$

I started with the base step:
$n(0) = 0^2+41(0)+41 = 41$
But I have no idea how to proceed in proving this.  Any tips or suggestions on getting start with this are appreciated.
 A: There is no non-constant polynomial in $\mathbb{Z}[x]$ that is able to take only prime values on $\mathbb{Z}$.
The reason is that:
$$ (b-a)\mid (p(b)-p(a)), $$
so, assuming that $p(a)=p_i$, then $p_i\mid p(a+kp_i)$ for every $k\geq 1$.
Assuming now that $p(a+kp_i)$ is a prime, it must be $\pm p_i$, so $p\pm p_i$ has an infinite number of integer roots, hence it is constant.
A: Yeah, if you let $n = 41$ then $(41)^2 + 41(41) + 41 = 3403$. Thus $3403$ can be factor in  two other numbers $41$ and $83$. Therefore this statement is not true for all $n \in Z$. 
A: Here is a slight alteration to Jack's response, suppose $p$ is a polynomial over $\mathbb Z$. Then if $p(a)$ is a multiple of $n$ so is $p(a+kn)$ for every integer $k$ since $a+kn\equiv a\bmod n$. Since a polynomial cannot take on the same value infinite times there must be a value say $b$ for which $p(b)$is a proper multiple of $n$. Using this we can prove any non-constant polynomial has an element with as many numbers in its prime factorization as is required. 
A: Clearly, for $n\equiv0 \mod(41)$, this term is product of two or three numbers, therefore not a prime.
