This from an exam I’m using for revision. It’s an unseen part and I’ve made an attempt at trying to answer it.
For bi) I’ve said that in a UFD, factorisations are only unique up until their units. Therefore, there are only finitely many factorisation of $k$ for which the product of $(x-y)(x+y)=k$. Therefore, there are only finitely many solutions with $x,y\in R$.
I don’t know whether that argument is correct. I have seen somewhere that this argument can be solved graphically by drawing a hyperbola, but could someone point me in the right direction with the attempt I’ve made, or if it’s wrong, give me some tips on how to progress?
You are essentially correct. Your argument is telling you, that there are only finitely many possibilities for $x+y$ and $x-y$. Then you just have to note that $$ x = \frac{1}{2}((x+y)+(x-y)), \qquad y = \frac{1}{2}((x+y)-(x-y)) $$
and hence, there are only finitely many possiblities for $x,y$.