How is the Pell-like equation $ax^2-by^2 = c$ solved?

Question

Suppose $a, b$ and $ab$ are non-square and solution exist for the above equation. One way we can solve it is by multiplying $a$ to the equation to get the Pell equation \begin{equation} (ax)^2-aby^2 = ac. \end{equation} If $(u,v)$ is the fundamental solution for the corresponding Pell equation and $(x_1,y_1)$ is a particular solution of the above equation, then all the solutions in the class of $(x_1,y_1)$ can be found by \begin{equation} (x_n+y_n\sqrt{ab}) = (x_1+y_1\sqrt{ab})(u+\sqrt{ab})^n. \end{equation} Are there any other way to find all the solutions if at all solution exist? Suppose $(u_0,v_0)$ is the fundamental solution of $ax^2-by^2 = 1$ and $(x_0,y_0)$ is a solution of $ax^2-by^2 = c,$ then I found that $(x_0\sqrt{a}+y_0\sqrt{a})(u_0\sqrt{a}+v_0\sqrt{b})^{2n}$ are also solutions but in the odd power I am getting numbers of the type $x'+y'\sqrt{ab}.$ How to obtain the solutions in this way?

Answer 1

You may also use this method:

$$ax^2-by^2=c$$

Suppose $x=y+d$ , putting this in equation we get:

$$(a-b)y^2+2ady+(ad^2-c)=0$$

We solve for y:

$$\Delta'=a^2d^2+(ad^2-c)(b-a)=ac+abd^2-bc$$

the condition for having solution in $\mathbb Z$ is:

$\Delta'=k^2$

In this case we have:

$$y=\frac{-ad\pm k}{a-b}$$

Only particular values of a, b, c give integer solutions. Some examples:

$$5x^2-7y^2=69$$

$\Delta'= k^2=(2-3)\times 69+2\times 3\times d^2$

which gives simple form of Pell like equation:

$$k^2-6d^2=-69$$

One solution to this equation is $(d, k)=(7, 15)$ which gives $(x, y)=(36, 29), (-1, 6)$.

$$23x^2-118y^2=5$$

Gives simple form $k^2-414d^2=25$

One solution to this equation is $(d, k)=(2, 41)$

which gives $(x, y)=(-1, 1), (-\frac {87}5, -\frac {77}5)$

From second solution we can derive an equation with integer solution:

$23(\frac{-77}5)^2-18(\frac{-87}5)^2=5$

$(x, y)=(\pm 77, \pm 87)$ will be the solution of this equation:

$23 x^2-18 y^2=125$