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!
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.