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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!

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  • $\begingroup$ Let $\alpha=2+3i$, $\beta=4+9i$. $\gcd(3,9)\ne1$, $\gcd(2,4)\ne1$. $\endgroup$ Feb 1, 2022 at 11:10
  • $\begingroup$ 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 $i$? $\endgroup$ Feb 1, 2022 at 11:12
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    $\begingroup$ Consider $14+15i$ and $10+21i$. $\endgroup$ Feb 1, 2022 at 11:15
  • $\begingroup$ It seems one odd integer divides the other odd integer is true but changing to the Gcd. That is interesting $\endgroup$
    – Liddo
    Feb 2, 2022 at 14:32
  • $\begingroup$ I don't understand your comment. Does the answer I posted answer your question? $\endgroup$ Feb 3, 2022 at 1:36

1 Answer 1

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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.

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