# What's the inverse element in the group of solutions to Pell's equation?

I'm working on a problem from a past exam paper,

Define the group operation on the set of solutions of Pell's equation, $$G_d=\{(x,y)\in\mathbb{Z}^2:x^2-dy^2=1\}$$ and show that the group axioms are satisfied.

I don't have access to course notes yet. I've surmised from this paper (adapted from their (1.2)) that the group operation required is $$(x_1,y_1)\star(x_2,y_2)=(x_1x_2+dy_1y_2,x_1y_2+x_2y_1).$$ Proving G1 and G2 is tedious but straightforward. I figured out G3. I so far haven't managed to figure out G4.

It's better to understand where $$\star$$ is from. Define $$\rho(x,y):=x + y \sqrt{d}$$ We have $$\rho((x_1, y_1)\star(x_2, y_2))=\rho(x_1, y_1)\rho(x_2, y_2).$$ Now everything follows naturally, since $$\{x+y\sqrt d\,|\,(x,y)\in G_d\}$$ forms a group under the usual multiplication of real numbers.
In particular, the inverse of $$(x,y)$$ under $$\star$$ is $$(x, -y)$$, as the inverse of $$x+\sqrt dy$$ under multiplication is $$x-\sqrt dy$$.
• I +1 it, but I think the answer would be better if you replaced "forms a group .." [which leaves one thinking one might have to check all the axioms] by something like "is a subgroup of the multiplicative group of non-zero real numbers, since it is clearly closed under multiplication, and the inverse of $x+y\sqrt{d}$ is ...". Jul 7 at 7:47