# Find positive integer solutions to $x^2-19y^2=1$

Find positive integer solutions to $x^2-19y^2=1$

I know that $(\pm13,\pm3)$ are odd-integer solutions to the generalized Pell: $x^2-19y^2=\color{brown}{-2}$, using this I have to find a solution for the Pell equation.

First Question: Does generalized Pell have either no solutions or infinitely many ? If so then I have to find another pair of solution $(x',y')$ such that $$(\pm13,\pm3)\neq(x',y')$$ $$x\equiv x'\mod r\ \text{and}\ y\equiv y'\mod r$$ then by a theorem If I set $(x'',y'')=\left(\frac{xx'-19yy'}{-2},\frac{-xy'+xy'}{-2}\right)$, then $\left(|x''|,|y''|\right)$ is a positive integer solution to the Pell equation, is that correct ?

$\bf{EDIT}:$ Using Will Jagy's formula I got $(x,y)=(170,39)$ as a solution, but where does that formula come from ?

• I would try to do $x^2 = t$ and $y^2=u$ so you have a diophantic equation. Commented Mar 18, 2015 at 16:00
• $(x=170,~y=39)~$ and $~(x=57799,~y=13260)$. Commented Mar 18, 2015 at 16:24
• not sure if this will help, $(x+4y)(x-4y) = 1+3y^2$
– JMP
Commented Mar 19, 2015 at 12:25
• @JonMarkPerry Thanks but, Is this a method to find directly the solution of the pell equation ? I think the intention of the exercise is to find another solution to the generalized pell s.t. the conditions above is satisfied and then derive the solution from there Commented Mar 19, 2015 at 13:01
• @inequal compute a table as in page 34 of the script of Andrew Kresch, but this time for $d=19$. You will find another solution $(x,y)$, which gives you $-2$ if you compute $x^2-19y^2$. Now you know that $x^2-dy^2 = N_{\Bbb{Q}(\sqrt{d})/\Bbb{Q}}$. Since the norm is multiplicative you see that $N_{\Bbb{Q}(\sqrt{d})/\Bbb{Q}}\left( \frac{13+3\sqrt{19}}{x+y\sqrt{19}}\right)=1$ and $\frac{13+3\sqrt{19}}{x+y\sqrt{19}}$ is exactely $170+39\sqrt{19}$ if you take $(x,y)$ to be the next solution after $(13,3)$ such that $x^2-dy^2 = -2$. :) Commented Mar 19, 2015 at 22:48

Regarding your "first question", given an initial solution to,

$$u^2-dv^2 = \pm2\tag1$$

then an infinite more can be found using the identity

$$(u x + d v y)^2 - d(u y + v x)^2 = (u^2-dv^2)(x^2-dy^2)$$

with the Pell equation,

$$x^2-dy^2=1\tag2$$

(Given an initial solution to $(2)$, I presume you know how to find all the others?)

Theorem (Legendre): The eqn $(1)$ is always solvable for,

1. $c=-2$, and prime $d=8n+3$,
2. $c=2$, and prime $d=8n+7$.

Given a $u,v$, then $(2)$ can be solved as,

$$(u^2\pm1)^2-d(uv)^2=1\tag3$$

For example, we have,

$$13^2-\color{brown}{19}\cdot3^2=-2$$ $$(13^2+1)^2-\color{brown}{19}\cdot(3\cdot13)^2=1$$

However,

$$\bigl(\bigl(39\sqrt{38}\bigr)^2+1\bigr)^2-\color{brown}{19}\cdot (39\cdot340)^2=1$$

showing that the relation $(3)$ does not yield all solutions to $(2)$.