How can I prove that a gaussian integer in the set A = $\mathbb{Z}[\sqrt{2}]$, is prime in A. Since I cannot necessarily use the property of it being congruent to 3 modulo 4.
In other word the set $A$ has elements of the form $a +b\sqrt{2}$ and I want to prove that a given number cannot be writen as a factor of two gaussian integer from that given set.
I will take the question to be, show that $5$ is not the product of two non-units of the form $a+b\sqrt2$ with $a,b$ integers.
So, suppose $5=\alpha\beta$ with neither $\alpha$ nor $\beta$ an integer.
Writing $\overline x$ for the conjugate of $x$, we get $25=\alpha\overline{\alpha}\beta\overline{\beta}$. Since neither $\alpha$ nor $\beta$ is a unit, we have $\alpha\overline\alpha=5$, so $a^2-2b^2=5$. Modulo $5$, this is $a^2\equiv2b^2\bmod5$. If $b\not\equiv0\bmod5$, then $c^2\equiv2\bmod5$, where $c\equiv ab^{-1}\bmod5$. But the only squares modulo $5$ are $0,1,4$, so this cannot be, so $b$ is a multiple of $5$. Then from $a^2-2b^2=5$ we see $a$ is also a multiple of $5$. But that makes the left side a multiple of $25$, which the right side isn't: contradiction! and we win.