Trying to solve a Pell equation (Diophantine equation $x^2 - ny^2 = 1$ where $n$ is not a square), we can generate all the solutions in the following way:

Given a fundamental solution $(x_1, y_1)$, we can find all other solutions by solving $${\displaystyle x_{k}+y_{k}{\sqrt {n}}=(x_{1}+y_{1}{\sqrt {n}})^{k},} \tag{*}\label{*} $$ for $k \in \mathbb{Z}$. (This can be solved by separating rational and irrational parts)

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My question is: why? That is, why would it be a good idea to look for solutions like this (i) , and more importantly, why does it give all solutions? (ii)

(Bonus question: same but for $x^2 + ny^2 = -1$ solvable)

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Anywhere that I could find this equation online, it is considered 'common' knowledge and therefore needs no explanation. I have not been able to find a proof.

What I have thus far: (i) $(x_1, y_1)$ is a solution, thus: $x_1^2 + ny_1^2 = 1$. Furthermore, for any $(x, y)$ we have $x^2 - ny^2 = (x - \sqrt{n}y)(x + \sqrt{n}y)$. Thus, for $k \in \mathbb{N}$ and any solution $(x, y)$:

$$ \begin{align} (x - \sqrt{n}y)(x + \sqrt{n}y) &= x^2 - ny^2 = 1 = 1^k\ \\ &= (x_1^2 - ny_1^2)^k = (x_1 - \sqrt{n}y_1)^k(x_1 + \sqrt{n}y_1)^k\end{align}$$ Now we can take only $(x + \sqrt{n}y)$ and $(x_1 + \sqrt{n}y_1)^k$ for some reason?

Also, if we assume \eqref{*} to be true, we can find the recurrence relations: $$\displaystyle x_{k+1} = x_1 x_k + n y_1 y_k,$$ $$\displaystyle y_{k+1} = x_1 y_k + y_1 x_k.$$ Since we can assume the fundamental solution won't have either zero I suppose we can conclude that each $x_k$ and $y_k$ will be unique. (ii) However, I still fail to see how this generates all solutions.


1 Answer 1


(i) Using (*) is simpler than calculating the convergents of $\sqrt n$, which is another way of solving Pell's equation.

(ii) This is Theorem 7.26 in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991:

Suppose that there is a positive solution $s, t$ that is not in the collection $\{x_k, y_k\}$. Because both $x_1 + y_1\sqrt n$ and $s + t\sqrt n$ are greater than 1, there must be some integer $m$ such that $$(x_1 + y_1\sqrt n)^m < s + t\sqrt n < (x_1 + y_1\sqrt n)^{m+1}.$$ Now $$(x_1 - y_1\sqrt n)^m = (x_1 + y_1\sqrt n)^{-m},$$ and we can multiply each part of the compound inequality by one or the other of those two equal expressions to obtain $$1 < (s + t\sqrt n)(x_1 - y_1\sqrt n)^m < x_1 + y_1\sqrt n.$$ Defining integers $a, b$ by $a + b\sqrt n = (s + t\sqrt n)(x_1 - y_1\sqrt n)^m$, we have $$a^2 - b^2n = (s^2 - t^2n)(x_1^2 - y_1^2n)^m = 1,$$ so $a,b$ is a solution of $x^2 - ny^2 = 1$ such that $1 < a + b\sqrt n < x_1 + y_1\sqrt n$. By the first part of that compound inequality and the fact that $0 < (a + b\sqrt n)^{-1} = a - b\sqrt n$, we have $0 < a - b\sqrt n < 1$.

Now \begin{align} a & = \frac{1}{2}(a + b\sqrt n) + \frac{1}{2}(a - b\sqrt n) > \frac{1}{2} + \, 0 \, > 0\\ b\sqrt n & = \frac{1}{2}(a + b\sqrt n) - \frac{1}{2}(a - b\sqrt n) > \frac{1}{2} - \frac{1}{2} = 0, \end{align} so $a, b$ is a positive solution. Therefore $a > x_1, b > y_1,$ which contradicts $a + b\sqrt n < x_1 + y_1\sqrt n$, and hence our supposition is false. Thus, all positive solutions are given by the collection $x_k, y_k$.

(Bonus) Now you should be able to prove the following statement, which is Problem 1 in Section 7.8 of An Introduction to the Theory of Numbers and comprises your bonus question:

Assume that $x^2 - ny^2 = -1$ is solvable. Let $x_1, y_1$ be the smallest solution. Then all solutions of $x^2 - ny^2 = -1$ are given by $x_k, y_k$ where $x_k + y_k\sqrt n = (x_1 + y_1\sqrt n)^k$ with $k = 1, 3, 5, 7, \ldots$, and that all solutions of $x^2 - ny^2 = 1$ are given by $x_k, y_k$ with $k = 2, 4, 6, 8, \ldots$.


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