4
$\begingroup$

Given an integer $n>8$, can we always find a prime $p$ such that

$$p = n^2-nb+b^2$$ where $b \in \{0,1,2,\ldots,n\}$?

For $n=8$ we have $n^2-nb+b^2 \in \{64,57,52,49,48,49,52,57,64\}$ which are all composite. But $n=8$ seems to be the largest $n$ where this does not hold. Is it possible to prove that there is always such a $b$ such that $n^2-nb+b^2$ is prime?

  • 1
    Do you know about [Eisenstein primes?](https://en.wikipedia.org/wiki/Eisenstein_prime) – 2017-02-07
  • 0
    Yes I do, this is actually where this question is coming from, but I wanted to see whether it might be easier to solve this in another context, (i.e. prime generating polynomials) – 2017-02-07

0 Answers 0