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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 ?

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  • $\begingroup$ I would try to do $x^2 = t$ and $y^2=u$ so you have a diophantic equation. $\endgroup$
    – Atvin
    Commented Mar 18, 2015 at 16:00
  • $\begingroup$ $(x=170,~y=39)~$ and $~(x=57799,~y=13260)$. $\endgroup$
    – Lucian
    Commented Mar 18, 2015 at 16:24
  • $\begingroup$ not sure if this will help, $(x+4y)(x-4y) = 1+3y^2$ $\endgroup$
    – JMP
    Commented Mar 19, 2015 at 12:25
  • $\begingroup$ @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 $\endgroup$
    – inequal
    Commented Mar 19, 2015 at 13:01
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    $\begingroup$ @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 $x^2-dy^2 = -2$. :) $\endgroup$
    – Bman72
    Commented Mar 19, 2015 at 22:48

1 Answer 1

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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 $(2)$.

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