Are Pell solutions "unique"??

Question

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 $j=k$?

Answer 1

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?

Answer 2

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.

Answer 3

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 $d \neq \pm 1.$