# Gaussian primes $\alpha$ and $\beta$ with $\mathrm{gcd}(\mathrm{Im}(\alpha),\mathrm{Im}(\beta))=1$

Let be $$F=\mathbb{Q}[i]$$. Also let $$\alpha=a+bi$$ and $$\beta=c+di$$ be primes in $$\mathbb{Z}[i]$$ such that $$N(\alpha), N(\beta)\equiv\;1\;(\mathrm{mod}\;4)$$ and $$N(\alpha)\not=N(\beta)$$

I am trying to see if it is true the next statement . "Multiplying by $$i$$ if it is necessary, we can assume without loss of generality that $$\mathrm{gcd}(\mathrm{Im}(\alpha),\mathrm{Im}(\beta))=1$$"

Since $$N(\alpha), N(\beta)\equiv\;1\;(\mathrm{mod}\;4)$$ and we that $$a$$ and $$b$$ have different parity and similar for $$c$$ and $$d$$ so I can assume that $$a$$ and $$c$$ are even integers and $$c$$ and $$d$$ are odd integers so I noticed by checking some Gaussian primes the following case:

"if $$b$$ divides $$d$$ then $$\mathrm{gcd}(a,d)=1$$ or $$\mathrm{gcd}(c,b)=1$$" which I proved by contradiction:

Since $$\mathrm{gcd}(a,d)\not=1$$ there is a primes $$q$$ such that $$q|a$$ and $$q|d$$. Similarly for $$\mathrm{gcd}(c,b)\not=1$$ we have a prime $$p$$ such that $$p|c$$ and $$p|b$$. Now since we have that $$b|d$$, it follows that $$p|d$$. Therefore $$N(\beta)=N(p)N(\tau)$$ where $$\tau\in \mathbb{Z}[i]$$ however we have a contradiction since $$N(\beta)$$ is a prime number.

If $$d$$ divides $$b$$ we hae something similar. But When I was checking some gaussian primes I found out cases where $$\mathrm{gcd}(b,d)\not=1$$ and in those case I found that $$\mathrm{gcd}(a,d)=1$$ or $$\mathrm{gcd}(c,b)=1$$. I could not prove it (if that is true in general) and there are some other case I missing out that I working on them.

Any hint or help would be great!

• Let $\alpha=2+3i$, $\beta=4+9i$. $\gcd(3,9)\ne1$, $\gcd(2,4)\ne1$. Feb 1, 2022 at 11:10
• Wait – when you say "multiplying by $i$", do you mean multiplying both $\alpha$ and $\beta$ by $i$, or do you just mean multiplying one of them by $i$? Feb 1, 2022 at 11:12
• Consider $14+15i$ and $10+21i$. Feb 1, 2022 at 11:15
• It seems one odd integer divides the other odd integer is true but changing to the Gcd. That is interesting Feb 2, 2022 at 14:32

The statement the question asks about is not true. Here's a counterexample.

Let $$\alpha=14+15i$$, and let $$\beta=10+21i$$.

Then the norm of $$\alpha$$ is $$14^2+15^2=421\equiv1\bmod4$$, and $$421$$ is a prime number, so $$\alpha$$ is a Gaussian prime.

The norm of $$\beta$$ is $$10^2+21^2=541\equiv1\bmod4$$, and $$541$$ is a prime number, so $$\beta$$ is a Gaussian prime. Also, $$421\ne541$$.

But the numbers $$\gcd(14,10)$$, $$\gcd(14,21)$$, $$\gcd(15,10)$$, and $$\gcd(15,21)$$ are all greater than one.