generalized way of finding pair solutions of an equation

Question

I want to find out pair solutions of this equation: $$x^{2}-79y^{2}=1$$ This is a hyperbola equation. I sketched its graph, but that didn't help me. I think the square from (form?) of $x$ and $y$ is the main problem for me. I haven't solved this type of pair solution before. There are two variables but only one equation. That's pretty weird to me. Is it really possible to solve? If yes, then how can I proceed? I want to know a generalized way to solve this type of problem.

Answer 1

The Pell equation $$ x^2 - Ny^2 = 1$$ (where $N$ is not a square) always has an infinite number of integer solutions, since there infinite units in the ring $\mathbb{Z}[\sqrt{N}]$. The "least" non-trivial solution can be found by expanding $\sqrt{N}$ into a continued fraction: the Legendre's theorem grants that such an expansion is periodic and the least element of the period is just $2\lfloor\sqrt{N}\rfloor$. In our case, since: $$\sqrt{79}=[8;1,7,1,16,1,7,1,16,1,7,1,16,\ldots],$$ we have that: $$\frac{80}{9}=[8;1,7,1]$$ is associated with the minimal solution: $$ 80^2-79\cdot 9^2 = 1,$$ and all the other non-trivial solutions can be found by expanding $$ (80+9\sqrt{79})^m$$ as an element of $\mathbb{Z}[\sqrt{79}]$.

For example, $(12799,1440)$ is a solution since $12799+1440\sqrt{79}=(80+9\sqrt{79})^2$.

We should have expected that, since $79=M^2-2$ with $M=9$, and the identity: $$ (M^2-1)^2 - (M^2-2)M^2 = 1$$ obviously holds.

Answer 2

There are infinitely many solutions $(x,y)$ to your problem, all of them must satisfy $$x^2-79y^2=1 \iff x^2 = 1+79y^2 \iff |x| = \sqrt{1+79y^2} \\\iff x = \sqrt{1+79y^2} \quad \text{ or } \quad x = -\sqrt{1+79y^2} .$$ And so you can write the set $S$ of all the solutions $(x,y)$ as follow $$S = \left\{(\sqrt{1+79t^2},t): t \in \mathbb{R}\right\} \cup \left\{(-\sqrt{1+79t^2},t): t \in \mathbb{R}\right\}$$

Answer 3

The first nontrivial solution to $x^2 - 79 y^2 = 1$ is $$ x=80, \; \; y = 9 $$

Given any solution $(x,y)$ you get another solution from $$ (80x + 711 y, 9x + 80 y). $$

$$ (1,0) $$ $$ (80,9) $$ $$ (12799,1440) $$ $$ (2047760,230391) $$ $$ (327628801,36861120) $$

Both $x$ and $y$ satisfy degree-two recurrences, $$ x_{n+2} = 160 x_{n+1} - x_n, $$ $$ y_{n+2} = 160 y_{n+1} - y_n. $$

For instance, $$ 12799 = 160 \cdot 80 - 1, $$ $$ 1440 = 160 \cdot 9 - 0. $$