Prime inheritance in class number 1 quadratic polynomials

Question

This conjecture is based on computational exploration of quadratic polynomials associated with imaginary quadratic fields of class number one.

Let us define: • For a polynomial $f(x) \in \mathbb{Z}[x]$, let $S = { f(0), f(1), f(2), \dots }$ be the sequence of integer outputs. • Suppose $f(n)$ is a composite number. • We say that $f$ satisfies the prime inheritance property if:

Every composite $f(n)$ has at least one prime factor that appeared in some earlier $f(k)$, for $k < n$, either as: • a prime value of $f$, or • a prime factor of a previous composite value of $f$.

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Example:

Consider $f(x) = x^2 + x + 41$. It is well-known that: • $f(x)$ is prime for $0 \le x < 40$, • and produces composites like $f(40) = 41^2 = 1681$, $f(41) = 41 \cdot 43$, $f(42) = 43 \cdot 47$, etc.

In all such cases (tested up to $x = 10^5$), every composite $f(n)$ has at least one prime factor that appeared in some $f(k)$ for $k < n$ — either as a prime output or a prime factor of an earlier composite.

This pattern appears to persist in the following polynomials:

$$ f(x) = x^2 + x + k \quad \text{where} \quad k \in {1, 2, 3, 5, 11, 17, 41} $$

Each corresponds to an imaginary quadratic field $\mathbb{Q}(\sqrt{1 - 4k})$ of class number 1.

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Conjecture (Prime Inheritance for Class Number 1):

Let $f(x) = x^2 + x + k$, where $\mathbb{Q}(\sqrt{1 - 4k})$ is an imaginary quadratic field of class number 1. Then for all $x \ge 0$, any composite value $f(x)$ has at least one prime factor that appears in a previous value $f(k)$ for $k < x$, either as a prime output or as a factor of a previous composite.

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Counterexamples in Higher Class Number:

For example, in $f(x) = x^2 + x + 4$, corresponding to $\mathbb{Q}(\sqrt{-15})$, we observe: • The first composite appears at $x = 3$: $f(3) = 3^2 + 3 + 4 = 16 = 2^4$ • But $f(0) = 4$, $f(1) = 6$, $f(2) = 10$ — and $2$ was already present

So the property still holds here initially, but further testing revealed violations. Thus:

The inheritance rule does not universally hold for all quadratic polynomials, but seems to be special to class number 1 fields.

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Questions: 1. Is this inheritance property known or previously studied? 2. Does the behavior of prime factor “propagation” in these polynomials reflect the deeper structure of the class group? 3. Is there a connection to how primes split in these number fields? 4. Can this be formalized or disproven for higher-degree or non-monic polynomials?

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I have tested this numerically up to $x = 100{,}000$ across all 9 class number 1 quadratic fields. All results consistently uphold the prime inheritance property (excluding trivial cases such as $f(0) = 1$).

Any insights, references, or historical context would be greatly appreciated.

Answer 1

I claim that if $f(n)$ has a prime divisor $q$ having size less than $n$, then $n$ is not the smallest positive integer $m$ such that $f(m) \equiv 0 \pmod{q}$. Thus, $f(n)$ "inherits" $q$ from an earlier term. Indeed, since $q | f(m)$ for some integer $m$, the congruence $f(x) \equiv 0 \pmod{q}$ is soluble, and in particular must have a solution in $\mathbb{Z}/q \mathbb{Z}$; that is, there exists $0 \leq m \leq q-1$ such that $f(m) \equiv 0 \pmod{q}$. Since $q < n$, the claim holds.

If $f(n)$ has at least three distinct prime divisors, then $f(n)$ must inherit the smallest of them from an earlier term in the sequence. Thus we focus on the case when $f(n)$ has exactly two distinct prime divisors (if $f(n)$ is itself prime, then clearly it cannot inherit anything). Thus we are left with the situation when every prime divisor of $f(n)$ is greater than $n$. There are only two ways this can happen: either $f(n) = q_1 q_2$ with $n < q_1 < q_2$ distinct primes, or $f(n) = q^2$ for some prime $q$.

We cover both scenarios at once, by insisting that $q_2 \geq q_1$. Since $q_2 \geq q_1 > n$, we must have $a,b \in \mathbb{Z}_{\geq 0}$ such that $q_1 = n + a, q_2 = n + b$. Then we have

$$\displaystyle n^2 + n(a+b) + ab = n^2 + n + k,$$

which implies that $n (a + b - 1) = k - ab$. Since the left hand side is non-negative, so must the right hand side; this implies that $ab \leq k$, and there are only $O_k(1)$ possibilities for $a,b$. This in turn shows that there are only finitely many options for $n$, bounded only in terms of $k$.

Since there are only finitely many $k = (p+1)/4$ such that $K = \mathbb{Q}(\sqrt{-p})$ has class number one, an explicit calculation can be carried out to verify that the property holds. However, I don't think class number one is necessarily the bottleneck.

Take $p = 163, k = 41$, which is your example. In this case we have $n(a+b-1) = 41 - ab$ implies that

$$\displaystyle n = \frac{41 - ab}{a + b - 1},$$

and in particular, we must have $n \leq 40$. Since you checked up to $n = 10^5$, there cannot be any counterexamples.

I think the reason why you are more likely to see this phenomenon with class number one is because almost by definition, these fields have no small split primes. I would imagine that the same thing could happen in quadratic fields where there are only a tiny number of small split primes, but I would be delighted to see a proof that this prime inheritance property always fails when the class number exceeds one.

Answer 2

The same reasoning by Daniel Weber works here as in your other post. Let $f(x)=x^2+x+k$. If $k<x$ then $f(x)<(x+1)^2$. So if $f(x)$ is composite then it has a prime factor $p\leqslant x$. Then $(x-p)^2+(x-p)+k$ is also divisible by $p$. I can't see anything special here about class number one.

Incidentally, in your conjecture, $k$ appears to have two different meanings, so it's hard to ascertain quite what you mean.