What are prime number values of the trinomial $q(n) = n^2 + n + 41$? Assuming $n$ is a positive integer

Question

Are there infinitely many integer values $n$ such that $q(n)$ is a prime number? Numerical evidence points to a yes answer.

This is similar to Landau's 4th problem from 1912. (The conjecture that there are infinitely many primes $p$ of the form $p=n^2+1 $?) Of course, Landau did not have a computer. Given n a positive number, for what values of $n$ is $q(n)=n^2 + n + 41 $ a prime number? This is known as Prime-Generating Polynomial.

see link https://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

also Wikipedia

https://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions

also my document on this quadratic polynomial

https://mersenneforum.org/showthread.php?p=581027#post581027

There are 3 .pdf files hosted at mersenne.org. I characterize all the cases when n^2 + n + 41 can be a composite number. Assuming positive integer n.

I show a data table, graph, and curve fit to characterize all the cases when this trinomial is a composite number, up to a certain numerical limit.

Also, I have found some algebraic factorizations for q(n)

https://sites.google.com/site/mattc1anderson/prime-producing-polynomial

A leading question is, "If we can know whenever n^2+n+41 is composite, what does that tell up about when that trinomial is a prime number?"

Let me know if there are any questions.

Matt

Answer 1

Since the ring of integers of $\mathbb{Q}[\sqrt{-163}]$ is a PID, it follows that a rational prime $p \neq 163$ may be expressed in the form $x^{2} + xy + 41y^{2}$ for rational integers $x$ and $y$ if and only if $p$ is a quadratic residue (mod $163$).(This is well-known). But, as you point out yourself, your question is comparable to asking how many primes $p$ have the form $n^{2} +1$ for integer $n$, which is well known to be open and Wojowu confirms in comments that your question is open too.

Later edit: I find it mildly interesting that the prime $p$ is expressible in this way (ie $p = n^{2}+n+41$) if and only if $p$ is expressible as the sum of four integer squares in one of the following ways: If $n$ is odd, we find that $p = \left( \frac{n-9}{2} \right)^{2} + \left( \frac{n+1}{2} \right)^{2} +\left( \frac{n+1}{2} \right)^{2} + \left( \frac{n+9}{2} \right)^{2}$ and if $n$ is even we find that $p = \left( \frac{n-8}{2} \right)^{2} + \left( \frac{n}{2} \right)^{2} +\left( \frac{n}{2} \right)^{2} + \left( \frac{n+10}{2} \right)^{2}.$

Even later edit: For any prime $p \neq 41$ which is a quadratic residue (mod $163$), there is a unique integer $h$ with $1 \leq h \leq \frac{p-1}{2}$ such that $p$ divides $h^{2}+h+41$, and then $p$ is necessarily the largest prime divisor of $h^{2}+h+41.$ An inductive argument of a type which dates back to Euler and/or Fermat then shows that $p$ is necessarily of the form $x^{2}+xy+41y^{2}$ for integers $x$ and $y$, and allows you to explicitly determine $x$ and $y$, given such an expression for the other (smaller) prime divisors $q$ of $h^{2}+h+41$ (all of which are also necessarily quadratic residues (mod $163$)).