Prime squares of the form $a^2+b^4$

Question

The Friedlander–Iwaniec theorem states there are infinitely many primes of the form $p=a^2+b^4$.

I am interested in whether there are infinitely many primes satisfying $p^2=a^2+b^4$ and $a\neq p$.

I tested Friedlander-Iwaniec primes up to 50000 for this property, and found 5, 41, 1201, 8521, and 41761. I did this in the naive hope of finding some pattern that might allow me to leverage the fact that $a^2+b^2=p \Rightarrow (a^2-b^2)^2+(2ab)^2=p^2$. However, I could not extract anything useful about these numbers in comparison to other Friedlander-Iwaniec primes to make progress.

Is this known? Would the sieve methods used in proving the Friedlander-Iwaniec theorem have any bearing on this question? (I have zero knowledge of these tools, so please excuse the naivete of my question) Or is this likely completely out of reach currently, as e.g. Landau's fourth problem is?

Answer 1

Not a solution. But we can reduce the problem to representation of the primes. We can apply Pythagorean triple formulas. $$p = x^2 + y^2 \quad a = 2xy \quad b^2 = x^2 - y^2$$ or $$p = x^2 + y^2 \quad a = x^2 - y^2 \quad b^2 = 2xy$$ In the first case we have $x - y = n^2$, $x + y = m^2$, $2p = m^4 + n^4$. $$p^2 = \left(\frac{m^4 - n^4}{2}\right)^2 + (mn)^4$$ In the second case we have $x = m^2$, $y = 2n^2$, $p = m^4 + 4n^4$. $$p^2 = \left(m^4 - 4n^4\right)^2 + (2mn)^4$$ So the problem was reduced to proving that there is infinitely many primes of the form $m^4 + 4n^4$ or $\frac{m^4 + n^4}{2}$.

Answer 2

$p^2=a^2+b^4\iff (a,b^2,p)$ is a Pythagorean triple so we have $$a=x^2-y^2\\b^2=2xy\\p=x^2+y^2$$ or $$a=2xy\\b^2=x^2-y^2\\p=x^2+y^2$$ (In both cases we can say the prime $p$ should be congruent to $1$ modulo $4$).

A simple possibility to have infinitely many primes $p$ as required, is to find an identity. For the first parameterization above we get $(4x^4-y^4)^2+(4x^2y^2)^2=(4x^4+y^4)^2$ which does not give many primes because $4x^4+y^4=(x^2-2xy+y^2)(x^2+2xy+y^2)$.

So it remains only the second parameterization to have a possibility of proof this way. From $b^2=x^2-y^2$ we get $$(b,y,x)=(r^2-s^2,2rs,x=r^2+s^2)$$ It follows the identity $$(4rs(r^2+s^2))^2+((r^2+s^2)^2-(2rs)^2)^2=((r^2+s^2)^2+(2rs)^2)^2$$ Are there infinite number of primes of the form $p=(r^2+s^2)^2+(2rs)^2=r^4+6r^2s^2+s^4?$

The equation $t^4+6t^2+1=0$ has as roots $t=\pm i\sqrt{3\pm2\sqrt2}$ so the polynomial $r^4+6r^2s^2+s^4$ is irreducible over $\mathbb Q$ (it is not so over $\mathbb Q(\sqrt2))$. It follows that for infinitely many rational integers $n$ the univalued polynomial $f_n(t)=t^4+6n^2t^2+n^4$ is irreducible over $\mathbb Q$ (and don't have a constant factor for each value of $t$).

According to an old Bunyakovsky's conjecture still unproved, for each $n$ there are infinitely many integers $k$ such that $f_n(k)$ is prime. Since for each value of $n$ we certainly can find a prime with $f_n(k)$ (if not, then it would be deny Bunyakovsky) we can assure that it is very probable that wonder by the O.P. is true.