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


also my document on this quadratic polynomial


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)


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.


  • 14
    $\begingroup$ This is an open problem. $\endgroup$
    – Wojowu
    Jun 15, 2021 at 9:17
  • 4
    $\begingroup$ How can numerical evidence point to something in this case? $\endgroup$
    – markvs
    Jun 15, 2021 at 13:15
  • 9
    $\begingroup$ The Bunyakovsky conjecture en.wikipedia.org/wiki/Bunyakovsky_conjecture says that, if $f(n)$ is an irreducible polynomial with positive leading term and there is no modulus $M$ for which $f(n)$ is identically $0 \bmod M$, then $f(n)$ is prime infinitely often. The polynomial $n^2+n+41$ satisfies these criteria. But there is no polynomial of degree $\geq 2$ for which the Bunyakovsky conjecture has been proved. $\endgroup$ Jun 15, 2021 at 13:44
  • 4
    $\begingroup$ @MarkSapir One can make a conjectural asymptotic for the number of $n< X$ with $n^2+n+41$ prime, find numerical evidence that this asymptotic holds, and observe that this asymptotic predicts that there are infinitely many prime values. $\endgroup$
    – Will Sawin
    Jun 21, 2021 at 0:37
  • 3
    $\begingroup$ Furthermore the method I suggested in this case, of first make a prediction based on the best available heuristics, then use numerics to check how well the heuristics seem to hold in this case, would not run into trouble in the 4n+1 vs. 4n+3 case anyways. $\endgroup$
    – Will Sawin
    Jun 28, 2021 at 18:40

1 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 $163$)).


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