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.