While reading this paper and after some experiments, I asked myself the following question.
Given
- the Gaussian integer $z_0=a+bi$, with $a\gt b$ and $b$ not divisible by $3$
- its conjugate $\bar{z_0}=a-bi$
- the function $w(z)=z^2+i$,
prove or disprove that if both $w(z_0)$ and $w(\bar{z_0})$ are Gaussian primes then, necessarily, $3|a$.
In other words, both $$(a^2+b^2)^2+4ab+1$$ $$(a^2+b^2)^2-4ab+1$$ have to be ordinary primes congruent to $1\;mod\;4$.
Some examples: $$z_0=3+i$$ $$z_0=6+4i$$ $$z_0=9+i$$ $$z_0=12+4i$$ $$z_0=15+5i$$
Many thanks.
Suppose that $3 \nmid a$. Then both $a$ and $b$ are $\equiv \pm 1 \pmod 3$, so
$$(a^2+b^2)^2 \equiv 1 \pmod 3, \\ 4ab \equiv \pm 1 \pmod 3.$$
There is some choice of sign in front of $4ab$ which makes that term $1 \pmod 3$, which makes $(a^2+b^2)^2 \pm 4ab + 1 \equiv 1 + 1 + 1 \equiv 0 \pmod 3$. It is impossible for this to be a prime congruent to $1$ mod $4$. This contradiction forces $3 \mid a$ as conjectured.