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Conjecture:
$\{a+b|a,b\in\mathbb N^+\wedge ma^2+nb^2\in\mathbb P^{>2}\}=\{k>2|\gcd(k,m+n)=1\}$ if $m,n\in \mathbb N^+$ and $\gcd(m,n)=1$.

This is a generalization of Any odd number is of form $a+b$ where $a^2+b^2$ is prime. Perhaps the generalization will spread some light of what is going on?

There is a perfect match of the formula for all tests I've done.

https://mathoverflow.net/questions/280123/the-set-of-numbers-ab-such-that-ma2nb2-is-prime

Lehs
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2 Answers2

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No. $$ m=1,n=91, a+b \neq 3 $$ although $$ \gcd(1,91) = 1, $$ $$ \gcd(3,92) = 1. $$ $$ 2^2 + 91 \cdot 1^2 = 95 $$ $$ 1^2 + 91 \cdot 2^2 = 365 $$

Looking at what I did, taking $m=1,$ we can take any $n \equiv 1 \pmod {30}$ with the same outcome. So, with $n=31,$ our two numbers that are not prime are $35$ and $125.$

Will Jagy
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Partial answer ...

Noting $A=\{a+b \mid a,b\in \mathbb{N}^{+} \wedge ma^2+nb^2 \in \mathbb{P}^{>2}\}$ and $B=\{k > 1 \mid \gcd(k,m+n)=1\}$ it's easy to show $A \subset B$.

  1. $a,b\in \mathbb{N}^{+} \Rightarrow a+b > 1$
  2. Now we need to show that $\gcd(a+b,m+n)=1$. Let's assume contrary $\gcd(a+b,m+n)=d>1$ then $a \equiv -b \pmod{d}$ and $m \equiv -n \pmod{d}$. Or $a^2 \equiv b^2 \pmod{d}$ and $m \equiv -n \pmod{d}$. Or $ma^2 \equiv -nb^2 \pmod{d} \Rightarrow d \mid ma^2 +nb^2$, so $d$ divides a prime $>2$ and $d \ne ma^2 +nb^2$ since $a+b\leq ma^2 +nb^2=d\leq a+b$, leading to $a+b=ma^2 +nb^2=2 \notin A$, a contradiction, thus $d=1$.
rtybase
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