# The set of numbers $a+b$ such that $ma^2+nb^2$ is prime

Conjecture:

If $m,n$ are coprime it exist a minimal natural number $N_{mn}$ such that:

$\{a+b>N_{mn}\mid a,b\in\mathbb N^+\wedge ma^2+nb^2\in\mathbb P_{>2}\} = \{ k > N_{mn} \mid \gcd(k,m+n)=1\}$.

Test for $k\leq 10^8$ suggest that $N_{11}=1$. The table below shows the largest minimal $N_{mn}$ for $k\leq 10^5$. In the table 0 stands for $\gcd(m,n)>1$.

    1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
1  1  1  1  1  1  4  1  1  1  7  1  3  1  1  1  4  1  5  1  1 13  1  1 17  1
2  1  0  1  0  1  0  4  0  1  0  1  0  1  0 31  0 18  0 20  0 10  0  1  0  1
3  1  1  0  1  1  0  1 19  0 10  1  0  3 11  0  6  7  0  1  1  0  9  3  0  1
4  1  0  1  0  8  0  1  0  4  0  1  0  5  0  1  0  5  0 14  0  6  0  1  0  6
5  1  1  1  8  0 16  1 21  1  0  1  1  1  1  0  1  1 22  1  0  1  8  1 18  0
6  4  0  0  0 16  0 20  0  0  0 38  0  1  0  0  0  1  0  6  0  0  0  5  0 16
7  1  4  1  1  1 20  0  1  1  1  1 14  1  0  1  1  1 12  1 19  0 15  1 12  1
8  1  0 19  0 21  0  1  0 10  0  1  0  4  0 19  0  9  0  5  0 13  0 18  0  8
9  1  1  0  4  1  0  1 10  0 17  1  0  1 14  0  1  1  0  3  5  0  6  1  0  1
10  7  0 10  0  0  0  1  0 17  0  1  0  1  0  0  0 11  0  1  0 10  0 28  0  0
11  1  1  1  1  1 38  1  1  1  1  0 11  1 16  1  4  1 12  1  6  3  0  1 29  1
12  3  0  0  0  1  0 14  0  0  0 11  0  9  0  0  0 18  0  1  0  0  0 24  0 16
13  1  1  3  5  1  1  1  4  1  1  1  9  0 10  1 20  1 23  1  1  1  1  5  1  1
14  1  0 11  0  1  0  0  0 14  0 16  0 10  0  1  0 33  0 37  0  0  0  1  0  1
15  1 31  0  1  0  0  1 19  0  0  1  0  1  1  0  6  1  0  1  0  0  1  1  0  0
16  4  0  6  0  1  0  1  0  1  0  4  0 20  0  6  0  5  0  9  0 16  0  1  0  1
17  1 18  7  5  1  1  1  9  1 11  1 18  1 33  1  5  0 16  1  7  1 20  1  1  1
18  5  0  0  0 22  0 12  0  0  0 12  0 23  0  0  0 16  0 22  0  0  0 23  0 11
19  1 20  1 14  1  6  1  5  3  1  1  1  1 37  1  9  1 22  0  8  1  5  1 18  1
20  1  0  1  0  0  0 19  0  5  0  6  0  1  0  0  0  7  0  8  0 17  0  8  0  0
21 13 10  0  6  1  0  0 13  0 10  3  0  1  0  0 16  1  0  1 17  0  1  1  0  1
22  1  0  9  0  8  0 15  0  6  0  0  0  1  0  1  0 20  0  5  0  1  0  7  0  1
23  1  1  3  1  1  5  1 18  1 28  1 24  5  1  1  1  1 23  1  8  1  7  0 16  1
24 17  0  0  0 18  0 12  0  0  0 29  0  1  0  0  0  1  0 18  0  0  0 16  0 10
25  1  1  1  6  0 16  1  8  1  0  1 16  1  1  0  1  1 11  1  0  1  1  1 10  0


My question is: are there heuristic reasons for or against the conjecture?

• Your conjecture follows from Schinzel's hypothesis, cf en.wikipedia.org/wiki/Schinzel%27s_hypothesis_H. Indeed, apply it to the irreducible polynomial $f_k(X) = m X^2 + n (k-X)^2$. The condition in Schinzel's hypothesis holds iff $k$ is prime to $m+n$ (assuming $m$ and $n$ are coprime). – js21 Sep 1 '17 at 13:24
• @js21 you have dropped the positivity condition on $a$ and $b$. – Will Sawin Sep 1 '17 at 13:35
• Oh, right, I missed it. – js21 Sep 1 '17 at 14:30
• I think @js21 is on the right track. The Schinzel hypothesis is not enough and neither is the Bateman-Horn conjecture but the heuristic pointing to them also suggests that $f_k(x)$ should be prime for some $x, 0 < x < k$ as soon as there is no local obstruction and $k$ is large enough. – Felipe Voloch Sep 2 '17 at 23:07