# Proving there are infinetly many integer solutions to $x^2 - 3y^2 = 1$

I am trying to show that there are infinitely many solutions to the following diophantine equation:

$$x^2 - 3y^2 = 1$$

But I don't really know where to start. I hear there are numerical ways to solve it, but it isn't the right direction. I also tried to use Minkowski's convex body theorem, but I can't use is for many disjoint sets since they have to be symmetric w.r. to 0

I'd appreciate any ideas

• This is the Pell's equation,this might help www-personal.umich.edu/~hlm/math475/pell.pdf – kingW3 Dec 10 '14 at 20:48
• There exist infinitely many units in $\mathbb{Z}[\sqrt{3}]$. Take all the powers of $2+\sqrt{3}$. – Crostul Dec 10 '14 at 20:48
• thanks, I actually aimed to prove the existence of infinitely many units in this ring. But now I know both algebraic and number-theory solutions – Jytug Dec 10 '14 at 21:10

Assume $x^2-3y^2=1$ for some integers x and y, then:

$$(2x + 3y)^2 - 3(x+2y)^2 = 4x^2 + 6xy + 9y^2 - 3x^2 - 6xy -12y^2=x^2 - 3y^2 = 1$$

And thus $(2x+3y, x+2y)$ is a new solution.

Since $(1, 0)$ is an integer solution and the above gives a new integer solution with larger x and y for each solution you find, there are infinitely many solutions.

A handy tool (which I also used to find the above) is Dario Alpern's generic two integer variable equation solver.

• This corresponds, in sardoj's answer, to the observation that if $x+y\sqrt 3$ has norm $1$, then $(x+y\sqrt 3)(2+\sqrt 3)$ also has norm $1$. – Mariano Suárez-Álvarez Dec 10 '14 at 21:14

Recall that in the ring $\mathbb{Z}[\sqrt{d}]$ where $d\in\mathbb{Z}$ is square-free, the norm of an element $x$ is given by $N(x)=x\bar{x}$ where $\bar{x}$ is the conjugate of $x$.

If we factor $x^2-3y^2=1$ we obtain

$$(x-\sqrt{3}y)(x+\sqrt{3}y)=1.$$

This is essentially showing us that the norm of $x+\sqrt{3}y$ is 1 (a unit). Also recall that the norm is multiplicative. That is, $$N(ab)=N(a)N(b).$$ Thus, if we can find one set of solutions to $x^2-3y^2=1$, say, $(x_0,y_0)$, then the pairs we get from computing $(x_0+\sqrt{3}y_0)^n, n\in\mathbb{N}$ will also be solutions. By inspection we can see that $(2,1)$ is a solution. Therefore, the pairs we get from $$(2+\sqrt{3})^n$$ will also be solutions. This shows that there are infinitely many solutions.