Simple Proof

I'm asked to prove that:

For any natural numbers a, b, c with c greater than or equal to 2, there is a natural number n such that an^2 + bn + c is not a prime.

I feel like my approach should be to let x be non-prime such that an^2 + bn + c = x. Then to show there is a solution n, in the natural numbers to an^2 + bn + (c - x) = 0, whilst somehow making use of the fact x can be wrote as the product of prime factors? Not sure where to get stuck in though..

Seems like quite a tricky gcse question

Proof by contradiction would be a good way to do this, suppose that there does not exist a natural number $n$ such that $an^2 + bn + c$ is not a prime. Then, only $1$ and $an^2 + bn + c$ divides $an^2 + bn + c$

However, since $c \geq 2$ the number $an^2 + bn + c > 2$ so it would be enough to show that there exists an $n$ such that $2$ divides $an^2 + bn + c$ - which is a contradiction.

ermm am i being a bit thick ? all you have to do is let n = c

how would i find the n? i swear we use a.b.c for this?

Just proceed by noting that if we take n=c then,

an^2 + bn + c = c(ac+b+1)

which is clearly not prime since divisible by c.