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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