0

Given the Pell equation $$X^2-2Y^2=1$$ and two solutions $(x_j,2uv)$ and $(x_k,2ac)$, with $uv=ac$ and $\gcd(u,v)=\gcd(a,c)=1$ and $u>v$ and $c>a$.

Can one prove that $(u,v)=(c,a)$, and hence $j=k$?

Kieren MacMillan
  • 7,889
  • 2
  • 28
  • 70
  • 2
    No. For example a solution is $X=17$, $Y=12$, and you could get this twice by taking $(u,v)=(6,1)$ and $(c,a)=(3,2)$. However since $Y=2uv=2ac$, it will certainly be true that $x_j=\pm x_k$. – David Jun 09 '14 at 13:41

3 Answers3

2

Don't really get the part about ac and so forth.

The equation $x^2 - n y^2 = 1$ is special in that all solutions lie in a single orbit of the automorphism group of the quadratic form $x^2 - n y^2.$ Find the "fundamental solution" $u^2 - n v^2 = 1$ with minimal $v > 0,$ and choose $u > 0.$ Then, every solution to $x^2 - n y^2$ with $x,y > 0$ is the left hand column of $A^k,$ where

$$ A \; = \; \left( \begin{array}{rr} u & nv \\ v & u \end{array} \right) . $$

Now, in general, the automorphism group is still infinite cyclic, for any indefinite binary quadratic form. However, there may be several distinct orbits for general $a x^2 + b x y + c y^2 = d,$ where $d \neq \pm 1.$

Will Jagy
  • 139,541
  • Here are all my cards on the table:

    For $X^2-2Y^2 = \pm 1$, I have found [algebraically] a solution $(F(u,v),2uv)$ and another $(c,a)$, with $u>v$ and $c>a$ and $uv=ac$, such that $(c,a)$ is a solution. By this last fact and a well-known identity, $(F(a,c),2ac)$ must also be a solution.

    What I'm trying to find out: Is this enough to conclude $(u,v)=(c,a)$?

     – Kieren MacMillan Jun 09 '14 at 23:19
  • Sorry, that comment implied something not necessarily true… What I meant was, $(c,a)$ being a solution to $X^2-2Y^2=\pm1$ implies that $(G(c,a),2ca)$ is also a solution [where $G(c,a)$ might or might not be distinct from $F(c,a)$, where $(F(u,v),2uv)$ is a solution].

    I guess what I'm asking is: Does $uv=ac$, with $u>v$ and $a>c$, plus the fact that $2ac$ and $2uv$ are both Pell numbers force $(u,v)=(c,a)$?

     – Kieren MacMillan Jun 09 '14 at 23:44
0

In general Pell's equation solutions are not unique because if $(x_0, y_0)$ is a solution to $x^2-Dy^2=1$, (where $D$ is a square free positive integer $>1$) then $$x_k+\sqrt{D}y_k =(x_0+\sqrt{D}y_0)^k$$ will be a solution to the same equation for all $k \geq 1$. Hence infinitely many solutions.

Anurag A
  • 41,067
0

It is clear now that the answer is "No". For example, $(17,12)$ is a solution to $X^2-2Y^2=1$. If $2ac=2uv=12$, we could have $(c,a)=(3,2)$ and $(u,v)=(6,1)$, as mentioned by David. Then one valid solution for the $17$ would be $17=3c+4a=2u+5v$; there are, in general, many others [because $\gcd(3,2)=\gcd(6,1)=1$].

Now the interesting question, I suppose, is: If $uv=ac$ but $(u,v)\ne(c,a)$, then what do the solutions around (i.e., "above" and "below") the common solution look like?

Kieren MacMillan
  • 7,889
  • 2
  • 28
  • 70