1

Background: If $P(n)$ is a reducible (over $\mathbb{Z})$ quadratic polynomial with integer coefficients, then all primes divide into at least one of the integers $P(n),\ n\in\mathbb{N}.$

However, for example, let $P(n)$ be the irreducible polynomial $P(n)=n^2+1.$ It is easily checked that $3,7,$ and $11$ do not divide into $P(n)$ for all integers $n.$ My conjecture is that in this example, there are infinitely many primes like the $3,7$ and $11.$

Conjecture: If $P(n)$ is an irreducible (over $\mathbb{Z})$ quadratic polynomial with integer coefficients, then there exist infinitely many primes that never divide into any of the integers $P(n),\ n\in\mathbb{N}.$

This is kind of the "other side" of https://en.wikipedia.org/wiki/Bunyakovsky_conjecture or https://en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H , which are about primes that do coincide with irreducible polynomials.

Is my conjecture true? If so, how do you prove it?

Adam Rubinson
  • 24,300
  • Sorry, I am a bit confusing about terminologies - the "divide into" part. So you are asking if for a given quadratic polynomial, there are infinitely many primes $p$ such that $p \nmid P(n)$ for every integer $n$, am I right? – Tri Dec 18 '24 at 02:34
  • For a given irreducible quadratic polynomial, yes. I don't think it's true if the quadratic is reducible. "Divide into" means, "is a factor of". Like $7$ divides into $14.$ – Adam Rubinson Dec 18 '24 at 02:38
  • Just one more clarification: By irreducible, you mean irreducible over $\mathbb{Q}$, right? I have an idea, so if it is what you mean, I will post it below. – Tri Dec 18 '24 at 02:58
  • This is likely a duplicate, but it follows easily from quadratic reciprocity and Dirichlet's theorem. – Eric Wofsey Dec 18 '24 at 03:00
  • No, I mean irreducible over $\mathbb{Z}.$ – Adam Rubinson Dec 18 '24 at 03:00
  • Any prime that does not divide the discriminant and for which $(\Delta |p) = -1$ – Will Jagy Dec 18 '24 at 03:00

2 Answers2

1

Too long for a comment: For irreducible over $\mathbb{Z}$, we can factor the greatest common divisor of $a,b,c$ and use Gauss lemma to reduce to irreducible over $\mathbb{Q}$.

I think this should be true. Let $P(x) =ax^2+bx+c$. Then $$4aP(x) = (2ax+b)^2-(b^2-4ac)$$ Since $P(x)$ is irreducible over $\mathbb{Q}$, the number $\Delta = b^2-4ac$ is not a square. We then have $$p \mid P(n) \Longrightarrow p \mid 4aP(n) \Longrightarrow \left(\dfrac{\Delta}{p}\right) =1,$$ where $\left(\dfrac{a}{p}\right)$ is the Legendre symbol. So if we can show that for a fixed integer $a$, there are infinitely many primes $p$ such that $a$ is a quadratic nonresidue modulo $p$, then we are done.

But this is always true. This is theorem 3 on page 57 in A Classical Introduction to Modern Number Theory By Kenneth Ireland, Michael Rosen. Or you can see here: Proving $(a/p) = -1$ for infinitely many $p$, without Dirichlet's Theorem.

Tri
  • 1,513
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 19 '24 at 22:16
  • @BillDubuque Sorry about the duplicated answer, I didn't mean to do that. It seems I was a bit careless when searching for the dupes. Should I delete it? – Tri Dec 19 '24 at 22:37
0

This holds for all irreducible polynomials of degree $>1$, see for example theorem 3.1.7 in The Chebotarev Density Theorem Applications, page 51:

Let $f(x)\in\mathbb{Z}[x]$ be an irreducible polynomial that has a zero modulo almost all primes $p$. Then $f(x)$ is linear.

Sil
  • 17,976