Let $f(n) = an^2 + bn + c$ be a quadratic function. Show that there's an $n ∈ N$ such that $f(n)$ is not a prime number. 
Let $f(n) = an^2 + bn + c$ be a quadratic function, where $a, b, c$
are natural numbers and $c ≥ 2$. Show that there is an $n ∈ N$ such
that $f(n)$ is not a prime number.

I figure this might have something to do with factorising the $f(n)$.
So, let $c=n$, then:
$f(n) = c(ac+b+1)$
The solutions say immediately after this step that $c(ac+b+1)$ is divisible by $c ≥ 2$, but I do not understand why or how this is so.
Any clarifications or further explanation on how to prove this would be much appreciated!
This is from a grade 11 Maths Specialist textbook under the topic of "for all" and "there exists" proofs.
 A: $f(n)$ is a function from the natural numbers to the natural numbers. Given $n$, we compute $f(n) = an^2+bn+c$. The question is to prove : there exists a natural number $n$ such that $f(n)$ is not prime. So, we must prove that one of $f(1),f(2),f(3),...$ is not prime.
Therefore, all we need is ONE value of $n$, or one natural number so that when I evaluate the function at that natural number, I get a composite number.
The proof then shows that $f(c)$ is not prime. Now, this mean that for ONE value of $n$, we have that $f(n)$ is not prime : that is, when we set $n=c$.

How is $f(c)$ shown to be not prime? We find a factor of $f(c)$ which is not $1$ or $f(c)$ : this shows that $f(c)$ is not prime.
By factorization, substituting $n=c$ gives $f(c) = ac^2+bc+c = c(ac+b+1)$, and $ac+b+1$ is a natural number, so $c$ divides $f(c)$, but $c \geq 2$ by assumption, and $ac+b+1 \geq ac \geq c \geq 2$, so $f(c)$ is not prime because there exists a factor of it which is not $1$ or itself(either $c$ or $ac+b+1$ will work).
Thus, we see that there exists an $n$ such that $f(n)$ is not prime.
A: Let $f(m)=p$, where $m\in\mathbb{N}$ and $p$ is prime. If $k\in\mathbb{N}$, then $f(m+kp)-f(m) \equiv 0 \pmod{p}$. If $f(m+kp)$ is a multiple of $p$ other than $p$, it is composite and we're done. Otherwise, $f(m+kp)=p \ \forall k$ which further implies that $f(n)=p \ \forall n$ (as we can always choose three or more distinct $k$'s) which cannot be the general case.
