I want to prove that prime elements in $\mathbb{Z}[\sqrt{2}]$ has one of the following forms:
(1) $\sqrt{2}$
(2) $p = 1,7 \quad mod \;\;8$
(3) integer primes $p=3,5 \;mod \; 8 $.
I was able to prove (1) and (3), but couldn't figure out (2).
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Suppose that $11$, let us say, we're reducible. Then we would render $11=(a+b\sqrt2)*(a-b\sqrt2)=a^2-2b^2$. If $b$ is even then $a^2-2b^2\in{0,1,4}\bmod8$, if $b$ is odd we get ${2,6,7}\bmod8$. There is a problem getting it to equal $11$, no? – Oscar Lanzi Feb 17 '22 at 00:29
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@OscarLanzi, why did you choose $11$? and why you assume that $11$ can be written as product of a number and its conjugate? – Adam_math Feb 17 '22 at 00:35
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2@Adam_math This link answers your question. Notice that (2) in your post is not a valid claim. – Kevin.S Feb 17 '22 at 02:13