Since gaussian primes can factor conventional prime numbers like 5, I was wondering about the same question but for quaternions.
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$3=(1+i+j)(1-i-j)$. We have $(i+j)^2=-2$ so this is quite like factoring $3$ in the ring $\Bbb{Z}[\sqrt{-2}]$. – Jyrki Lahtonen Aug 27 '21 at 07:09
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The number $11$ will do the job. It's not a sum of two squares, so it can't be a product of Gaussian primes, but it can be written $$ 11 = ( 1 + 3i + j)(1 - 3i - j). $$
hunter
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(If you want to learn more about this question and how to approach it systematically, you will want some tools from Algebraic Number Theory.) – hunter Aug 24 '21 at 18:47
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