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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.

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1… – 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
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 $(a,b)$

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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.

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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

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.

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