Determining if all integers of the polynomial form $n^2+21n+1$ are prime

Question

Suppose I had a statement that said For all positive integers of n, ${n^2 + 21n + 1}$ is prime.

Attempt:

The first thing that I decided to do was to try and factor it. I immediately saw that it wasn't going to happen. This meant that the GCF for each term was one.

I then checked using the discriminant method and my result was ${\sqrt{437}}$ Though 437 is not prime since it is divisible by 19 and 23

My question is, is this enough to say that for all integers of this polynomial, ${n^2 + 21n + 1}$ is prime. If not, how else can I efficiently check to see that their exists at least one case that makes this not true.

Answer 1

When $n=1$, our polynomial is $23$. Now evaluate it at $n=1+23$. From "failure," success!

Remark: The same basic idea can be used to show that no non-constant polynomial $P(n)$ with integer coefficients can be prime for all natural numbers $n$.

Answer 2

André Nicholas illustrates a general observation which is easy and which works more generally.

If you wanted to continue your own observation further, one way would be:

$$f(n)=n^2+21n+1$$

$$4f(n)=(2n+21)^2-437$$

So $f(n)$ would have a factor $437$ whenever $2n+21$ is a multiple of $437$.

And since $437=441-4=21^2-2^2=19\times 23$, $f(n)$ will have a factor $19$ whenever $2n+21$ is divisible by $19$, and $23$ whenever $2n+21$ is divisible by 23.

You just need to confirm that the relevant multiple is greater than $1\times 437, 19, 23$ most of the time.

Answer 3

I would say $n^2+21n+1=(n+1)^2+19n$, so if $n+1$ has a common factor with $19$, the expression will be divisible by $19$. In fact, $18^2+21\cdot 18+1=703=19\cdot 37$