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I want to show that $2$ and $2\sqrt{2}$ are irreducible in $\mathbb{Z}[2\sqrt{2}]$.

Consider the norm $N:\mathbb{Z}[2\sqrt{2}]\to\mathbb{Z}_{\ge0}$ defined by $N(a+b\cdot2\sqrt{2})=a^{2}-8b^{2}$.

Assume that $2=(a+b\cdot 2\sqrt{2})(c+d\cdot2\sqrt{2})$ for some $a,b,c,d\in\mathbb{Z}$.

Then $4=(a^{2}-8b^{2})(c^{2}-8d^{2}),$ by multiplicativity of $N$.

So, I tried to show that if one of the factors in RHS of the previous equation is $\pm2$, it leads to a contradiction.

Is it true that there is no integer solution to the equation $a^{2}-8b^{2}=\pm2$? How do I prove it?

I'm not sure that the congruence $a^{2}\equiv\pm2\pmod{8}$ has no solution.

Is it the right way to show the equation $a^{2}-8b^{2}=\pm2$ has no integer solution?

Give some advice. Thank you!

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    $\begingroup$ Try squaring $0,1,2,3,4,5,6$, and $7$ modulo $8$; you'll find the result is $0, 1, $ or $4$ $\endgroup$ – J. W. Tanner Apr 23 at 16:07
  • $\begingroup$ Answering your first question: $a^2=8b^2\pm 2$ so $a^2=2(4b^2\pm 1)$ and so $2$ divides $a^2$ and therefore $a$. Meaning $4$ divides $2(4b^2\pm 1)$ which cannot happen. $\endgroup$ – freakish Apr 23 at 16:10
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It is true that there is no integer solution to $a^2-8b^2=\pm2$,

and your idea of proving that by looking at this equation modulo $8$ is a good one.

(Modulo $4$ would work too.)

If $a$ is odd, then $a=2k+1$ so $a^2=4k^2+4k+1=4k(k+1)+1\equiv1\pmod 8$;

if $a$ is even, then $a=2k$ so $a^2=4k^2\equiv 0 $ or $4 \pmod 8$.

In any event, $a^2\equiv\pm2\pmod8$ has no solutions, so $a^2-8b^2=\pm2$ has no integer solutions.

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