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.

  • 3
    $\begingroup$ 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. $\endgroup$ – lulu Feb 22 '18 at 20:40
  • 1
    $\begingroup$ 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. $\endgroup$ – lulu Feb 22 '18 at 20:44
  • $\begingroup$ 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 $>1$). $\endgroup$ – lulu Feb 22 '18 at 20:47
  • $\begingroup$ Great, thanks for the link! Very helpful. $\endgroup$ – Samuel Feb 22 '18 at 20:49
  • $\begingroup$ @lulu I'm not quite sure how to understand your second comment. Why do $f(0)$ and $f(1)$ have to be divisible by $2$? $\endgroup$ – 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.

To stress: no affirmative examples of this are known. Worse, it is unknown whether such a polynomial must in fact take even a single prime value. There are examples where the first prime value occurs for unexpectedly high arguments.

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.

  • $\begingroup$ 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 $\endgroup$ – DanielWainfleet Feb 22 '18 at 23:22
  • $\begingroup$ @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. $\endgroup$ – lulu Feb 22 '18 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.