# Quadratics which produce no primes

There are some famous polynomials which produce a series of consecutive prime numbers, including Euler's $n^2 + n + 41$, which produces primes for $0 \leq n \leq 39$.

What I've been thinking about recently are quadratic polynomials with integer coefficients which don't produce any primes at all for $n \geq 0$. There are trivial cases:

$$n^2 + n$$

$$an^2 + bn + c \quad,\quad \gcd(a, b, c) > 1$$

There's at least one family of polynomials which is borderline trivial, but worth mentioning:

$$n^2 + (2a - 1)n + a^2 \quad,\quad a\quad \text{even}$$

All values of polynomials of this form are even; the reader can easily check using parity rules. What I am very curious about is the case when $a$ is odd.

Does there exist an $a \in \mathbb{N}$ such that $$f_a(n) = n^2 + (2a - 1)n + a^2 \quad,\quad a\quad \text{odd}$$ produces no primes for $n \geq 0$?

It can be shown using parity rules that $f_a$ is odd for all $n$.

Any thoughts appreciated.

• the Bouniakowsky Conjecture would tell us that every irreducible monic polynomial with integer coefficients is prime infinitely often unless there is a single prime dividing all the values at integers. – lulu Feb 22 '18 at 20:40
• In your case, we remark that $f(0)=a^2, f(-1)=a^2-2a+2$ so any prime $p$ which divided both $f(0),f(1)$ would have to divide $2$, whence we'd have to have $p=2$, which you have excluded. – lulu Feb 22 '18 at 20:44
• side note: on reflection, I don't think you need "monic" in the conjecture. It is, of course, all academic, as no affirmative examples are known (for degree $>1$). – lulu Feb 22 '18 at 20:47
• Great, thanks for the link! Very helpful. – Samuel Feb 22 '18 at 20:49
• @lulu I'm not quite sure how to understand your second comment. Why do $f(0)$ and $f(1)$ have to be divisible by $2$? – Samuel Feb 22 '18 at 21:10

Very little is known about questions of this form. Surprisingly little, really. The Bouniakowsky Conjecture would tell us that any irreducible polynomial (of degree $>1$) with integer coefficients takes infinitely many prime values UNLESS there is some prime $p$ which divides every one of the values of the polynomial at integers.
We remark that your polynomials satisfy the conditions of the conjecture. They are irreducible over $\mathbb R$, so certainly irreducible over $\mathbb Z$. And there is no prime that divides all the values at integers. Indeed, we compute: $$f(0)=a^2\quad \& \quad f(-1)=a^2-2a+2$$
Any prime $p$ which divided both of those would have to divide $2$ (as it would have to divide $a$). But $a$ is assumed to be odd so this is impossible.
• According to your link, one of the conditions of the B, conjecture is that no prime divides $f(n)$ for all $n\in \Bbb N$ (as opposed to all $n\in \Bbb Z).$... But it is easily shown that if prime $p$ divides $f(1),f(2),$ and $f(3)$ then $p=2,$ which is impossible as $f(n)$ is always odd................+1 – DanielWainfleet Feb 22 '18 at 23:22
• @DanielWainfleet The conditions ($\mathbb N$ or $\mathbb Z$) are basically the same...here, for instance, just look at $g(x)=f(x-2)$. $g(x)$ takes infinitely many prime values for natural number arguments iff $f(x)$ does, and for $g(x)$ my values suffice. – lulu Feb 22 '18 at 23:29