Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm interested in finding positive integers which satisfy an equation.

I've been thinking about the following equation: $$x^2-10y^2=1\ \ \ \ \ \ \cdots(\star).$$

Then, I've just got the following (let's call this theorem):

Theorem: If $(x,y)$ satisfies $(\star)$, then $(20y^2+1,2xy)$ also satisfies $(\star)$.

Proof: Letting $x=10n+1$, we get $2n(5n+1)=y^2$. Hence, let's consider the case in which both $2n$ and $5n+1$ are square numbers. Then, we can represent $n=2k^2$, so we get $5n+1=10k^2+1$. Hence, $y=k$ is sufficient because of $(\star)$.

Then, letting $n=2y^2$, then we get $$2n(5n+1)=2\times2y^2\times(10y^2+1)=(2xy)^2\ \Rightarrow\ (20y^2+1)^2-10(2xy)^2=1$$ Now the proof is completed.

It's easy to get $(x,y)=(19,6)$, so by using this theorem, we know we can get an infinite number of sets $(x,y)$ as the following: $$(19,6), (721,228), (1039681,328776),\cdots$$

By the way, by using computer, I found that $(x,y)=(27379,8658)$ also satisfies $(\star)$ though this set cannot be got from the theorem above.

Then, here is my question.

Question: How can we get all positive-integers sets $(x,y)$ which satisfies $(\star)$ ?

I've tried, but I'm facing difficulty. Any help would be appriciated.

share|improve this question
en.wikipedia.org/wiki/… –  oldrinb Aug 29 '13 at 12:35
Find an intro Number Theory textbook, go to the chapter on Pell equations, read up on it, then come back and post an answer to your own question. –  Gerry Myerson Aug 29 '13 at 12:36

4 Answers 4

up vote 2 down vote accepted

Note that if $a^2-10b^2=1$ we have $(a+\sqrt{10}b)(a-\sqrt{10}b)=1$ and also therefore that $$(a+\sqrt{10}b)^r(a-\sqrt{10}b)^r=1$$ So that $(a+\sqrt{10}b)^r=A+\sqrt{10}B$ generates a solution $(A,B)$

Also if we know that $(a,b)$ is a solution, and $(c,d)$ is a solution then


and we can take $(a+\sqrt{10}b)(c+\sqrt{10}d)=(ac+10bd)+(ad+bc)\sqrt{10}$ which yields the new solution $(ac+10bd, ad+bc)$ - and we can always use the smallest solution $(19,6)$ for $(a,b)$

share|improve this answer

The solutions $(x_n,y_n)$ can be easily computed recursively by hand. Set $(x_0,y_0)=(-1,0)$ or $(1,0)$. Then define $$ (x_{n+1},y_{n+1})=(19x_n+60y_n,6x_n+19y_n). $$ If you start with $(1,0)$, then the next one is the fundamental solution $(19,6)$. It is easy to verify that this yields all solutions.

share|improve this answer
How did you figure this out? –  Ataraxia Aug 29 '13 at 13:53
These are the so called coupled recurrence formulas for Pell's equation. They are well known. –  Dietrich Burde Aug 29 '13 at 14:02
@Ataraxia $(x_{n+1},y_{n+1})=(x_0 x_n+Dy_0 y_n,x_0 y_n+y_0 x_n)$, where $x^2-Dy^2=1$. –  Cortizol Aug 29 '13 at 14:07
@Ataraxia If you look at my solution, you might see a little more of where this comes from. –  Mark Bennet Aug 29 '13 at 14:14

This is an example of Pell's equation. To find all the solutions you first find a fundamental solution which is a convergent of the continued fraction for $\sqrt{10}$. Once you have this solution, you can generate all the others. See the solutions section for more details.

share|improve this answer

The equation $x^2-10y^2=-1$ has the obvious solution $x=3$, $y=1$. Using the theory of the Pell equation, one can show that the solutions of your equation are given by $$x_n=\frac{(3+\sqrt{10})^{2n}+(3-\sqrt{10})^{2n}}{2},\qquad y_n=\frac{(3+\sqrt{10})^{2n}-(3-\sqrt{10})^{2n}}{2\sqrt{10}}.$$

Remark: The Pell equation is discussed in most books in Elementary Number Theory. I also recommend the very nice book on the Pell equation by Ed Barbeau.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.