Primes of the form $x^2 +ny^2$ where swapping $x$ and $y$ still gives a prime

I am studying primes of the form $x^2+ny^2$, and i was wondering if there are any known results about the primes of this form such that when you swap $x$ and $y$ you also get a prime. ie for $y^2+nx^2$ you get another prime, I have found quite a lot of these primes and was wondering if there might be infinitely many of these pairs of primes.

For $n=2$ I have found for example: $11, 43, 59, 67, 83, 107, 139, 163, 179, 211, \ldots$, all of these primes can be written as $x^2+2y^2$ and for these values of $x$ and $y$, $y^2+2x^2$ is also a prime.

Thank you!

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I don't know the answer to your question, but I would guess that a good place to start would be David Cox's book Primes of the form $x^2+ny^2$. –  Arturo Magidin Jun 2 '11 at 19:53
Is your list complete? I found 3,11,19,43,59,67,83,107,139,163,179,211,251,307,331,419,443,... –  lhf Jun 3 '11 at 0:37
Thank you for the help I do have that book, you're right my list was missing some primes, and yeah i had noticed the 3 mod 8 relationship, but I'm more interested in the general case, I wondered if for any n there are infinitely many of these prime pairs, or if this statement is a result of a known conjecture. –  Chris Jun 3 '11 at 9:14
@Chris: please do not use answers to make comments. –  Qiaochu Yuan Jun 3 '11 at 10:13

This could make a nice trick question:

Prove or disprove if $p=x^2+2 y^2$ is a prime which is $3 \mod 8$, then $y^2+2x^2$ is prime.

The answer is "disprove". The first counter-example I can find is $p=131$, which is $9^2+2 \times 5^2$, where $5^2+2 \times 9^2 = 187=11 \times 17$. As I will explain below, there is a good reason where there are no significantly smaller counter-examples.

First of all, let me explain the $3 \mod 8$ business. An odd prime is of the form $x^2+2 y^2$ if and only if it is $1$ or $3 \mod 8$. If $p$ is $1 \mod 8$, then $y$ is even and $y^2+2x^2$ is not prime. So the only primes we need to care about are the ones which are $3 \mod 8$.

Let $z=y^2+2x^2$ and let's think about what a prime divisor $q$ of $z$ could look like. $z$ is odd, so $q$ isn't $2$. Also, $x^2+2y^2$ is divisible by $3$ if and only if $y^2+2 x^2$ is, so $q$ isn't $3$. Since $x^2+2y^2$ is prime, $x$ and $y$ are relatively prime. So $q$ divides neither $x$ nor $y$ and we can change $x^2 \equiv 2 y^2 \mod q$ into $(x/y)^2 \equiv -2 \mod q$. So $-2$ is a square modulo $q$, meaning that $q$ is $1$ or $3$ mod $8$.

To give a counter-example, $z$ must be a number which is $3 \mod 8$, not divisible by $3$, and has all prime factors equal to $1$ or $3 \mod 8$. The first such number is $187$ and, sure enough, that gives a counter-example. In fact, it gives two: $187$ is also $13^2+2 \times 3^2$, and $3^2+2 \times 13^2 = 347$, which is prime. Some other values worth trying are $17 \times 19 = 323$, $11 \times 41 = 451$ and $17 \times 43 = 731$.

Here is the experiment I would do if I wanted to pursue this further: Generate a collection of random pairs $(x,y)$ with $x$ and $y$ both odd, relatively prime, and exactly one of $(x,y)$ divisible by $3$. (Just generate random integers between, say $1$ and $10^4$ with the right properties modulo $2$ and $3$, and reject pairs which have a common factor.) Compute the proportion of pairs for which $x^2+2y^2$ is prime, call this proportion $p$, and the proportion of pairs for which $x^2+2 y^2$ and $y^2+2 x^2$ are both prime, call that proportion $q$.

I would bet $q$ is pretty close to $p^2$. If so, I think that the obvious explanation covers all there is to see here.

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