# How to show that some elements of $\mathbb{Z}[2\sqrt{2}]$ are irreducible?

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!

• Try squaring $0,1,2,3,4,5,6$, and $7$ modulo $8$; you'll find the result is $0, 1,$ or $4$ – J. W. Tanner Apr 23 at 16:07
• 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. – freakish Apr 23 at 16:10

## 1 Answer

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.