I'm working my way though a classical geometry book by Hartshorne right now, but this problem popped up in a section I'm reading. It is Problem 13.10 from Hartshorne's Geometry: Euclid and Beyond if you're curious.
Anyway, the problem states:
If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$, then the square root is actually in $\mathbb{Z}[\sqrt{2}]$.
I'm not super familiar with algebra, so I'm having trouble interpreting the question, but I would like to know how to solve it.
I looked up $\mathbb{Q}(\sqrt{2})$ on wikipedia, and it seems that it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Q}\}$. I couldn't find $\mathbb{Z}[\sqrt{2}]$, but I assume it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Z}\}$.
So if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$ means there exists some $c+d\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ such that $ (c+d\sqrt{2})^2=c^2+2d^2+2cd\sqrt{2}=a+b\sqrt{2}. $ This implies (I think?) that $c^2+2d^2=a$ and $2cd=b$. If this is the correct path, is there then someway to conclude that $c$ and $d$ are in fact integers? Thanks.
By the way, is this exercise easily related to some aspect of classical geometry? It seems kind of out of the blue to me.