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.
$f(n)$ 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.