# Solutions to Pell’s Equation

A Pell’s equation has one trivial solution, $$(x,y)=(1,0)$$, corresponding to solution $$z=1$$ of equation $$N(z)=1$$. But if we know the smallest non-trivial solution, then we can derive all the solutions. This is what the following statement claims.

Theorem 2

If $$z_0$$ is the minimal element of $$\mathbb{Z} [\sqrt{d}]$$ with $$z_0> 1$$ and $$N(z_0)=1$$, then all the elements $$z\in\mathbb{Z}[\sqrt{d}]$$ with $$Nz=1$$ are given by $$z=\pm z_0^n$$, $$n\in\mathbb{Z}$$.

Corollary

If $$(x_0,y_0)$$ is the smallest solution of the Pell’s equation with $$d$$ given, then all natural solutions $$(x,y)$$ of the equation are given by $$x+y\sqrt{d}=\pm(x_0+y_0\sqrt{d})^n$$, $$n\in\mathbb{N}$$.

Note that $$z=x+y\sqrt d$$ determines $$x$$ and $$y$$ by the formulas $$x=\frac{z+\overline{z}}2$$ and $$y=\frac{z-\overline{z}}{2\sqrt d}$$. Thus all the solutions of the Pell’s equation are given by the formulas $x=\frac{z_0^n+\overline{z_0}^n}2\quad\mbox{i}\quad y=\frac{z_0^n-\overline{z_0}^n}{2\sqrt d}.$

Example

The smallest non-trivial solution of the equation $$x^2-2y^2=1$$ is $$(x,y)=(3,2)$$. Therefore for every solution $$(x,y)$$ there is an integer $$n$$ such that $$x+y\sqrt d=\pm(3+ 2\sqrt2)^n$$. Thus $x=\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}2,\quad y=\frac{(3+2\sqrt2)^n-(3-2\sqrt2)^n}{2\sqrt2}.$

Now we will show that every Pell’s equation indeed has a non-trivial solution.

Lemma 1 (Dirichlet’s theorem)

Let $$\alpha$$ be an irrational number and $$n$$ be a positive integer. There exist $$p\in\mathbb{Z}$$ and $$q\in\{1,2,\dots,n\}$$ such that $$\left|\alpha-\frac pq\right|< \frac1{(n+1)q}$$.

Lemma 2

If $$\alpha$$ is an arbitrary real number, then there exist infinitely many pairs of positive integers $$(p,q)$$ satisfying $$\left|\alpha-\frac pq\right|< \frac1{q^2}$$.

Theorem 3

A Pell’s equation has a solution in the set of positive integers.

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