# Trouble understanding generalised form of Pell’s equation

A Pell’s equation is a diophantine equation in $$x$$ and $$y$$ of the form $$x^2-d\cdot y^2=1$$ with $$d$$ a square-free integer.

The fundamental solution of a Pell’s equation is the smallest (with the smallest absolute value) number $$z_1=x_1+y_1\cdot\sqrt{d}$$ for which $$(x,y)$$ is a solution to the Pell’s equation and $$|z_1|>1$$.

In this case, I know (and understand) that the general solution to the equation is $$z_n=z_1^n$$.

The general form of Pell’s equation is a diophantine equation of the form $$r\cdot x^2 - d\cdot y^2=c$$ for which $$r>0$$, $$d>0$$ and $$c\neq0$$ are integers.

And now, my book says that a (there can be multiple) fundamental solution to this generalised form is a number $$z_0=x_0+y_0\sqrt{d}$$ for which $$x_0^2-d\cdot y_0^2=c$$ and $$1<|x_0+y_0\sqrt{d}|. Here I assume that $$z_1$$ is the fundamental solution of the corresponding ‘simple’ Pell’s equation $$x^2-d\cdot y^2=1$$.

Next, it should be possible to calculate the general solution $$z_n=z_0\cdot z_1^n$$.

However, I don’t understand how to calculate a fundamental solution of the generalised equation and why the general term should equal $$z_0\cdot z_1^n$$.

For example, consider the diophantine equation $$x^2-2y^2=-1$$.

Then $$z_1=3+2\sqrt{2}$$, as $$(3,2)$$ is the fundamental solution of $$x^2-2y^2=1$$.

Now, I would expect that there are no fundamental solutions to the original equation, as there are no numbers $$z_0=x_0+y_0\sqrt{d}$$ for which $$(x_0,y_0)$$ is a solution to $$x^2-2y^2=-1$$ and $$1<|z_0|<|z_1|$$.

However, the original equation does have integer solutions! (take, for example, $$(x,y)=(7,5)$$

Could someone please explain me what I’m doing wrong?

• What are $z_0,z_1$ in inequality (after "and" of third last sentence) and why is that inequality imposed? – coffeemath Apr 2 at 19:25
• what book are you using? – Will Jagy Apr 2 at 20:12

In your example, $$1-\sqrt{2}$$ generates the solution $$x=1,y=-1$$ and this does satisy the fundamental solution inequality. The equivalent $$z_0 = \sqrt{2}-1$$ is easier to work with. With $$z_1 = 3+2\sqrt{2}$$, this $$z_0$$ generates the following solutions: $$z_0z_1 = 1+\sqrt{2} \Longrightarrow (x=1,y=1) \\ z_0z_1^2 = 7+5\sqrt{2} \Longrightarrow (x=7,y=5) \\ z_0z_1^3 = 41+29\sqrt{2} \Longrightarrow (x=41,y=29) \\ z_0z_1^3 = 239+169\sqrt{2} \Longrightarrow (x=239,y=169) \\ \vdots$$ Nothing cute tells me how to find that first $$z_0$$ in general, but that is the same situation as for the original Pell equation, where nothing tells you how to find the smallest $$z_1$$. In fact, the situation is better here because at least you know a limit on how large $$z_0$$ can be.
I don't know that an exhaustive search revealing no $$z_0$$ in the range $$1 \leq |x_0+y_0\sqrt{d}| < z_1$$ proves that the corresponding generalized Pell equation has no solution, but I suspect that is the case.