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 $1<|x_0+y_0\sqrt{d}|<z_1$. 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?

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

  • $\begingroup$ I think that z_0 should be 1+\sqrt{2}, because |z_0| must be larger than 1. But (at least in this example) it doesn’t matter, as it generates the same solutions as z_0 = \sqrt{2}-1. (This is because (\sqrt{2}-1)(3+2\sqrt{2}) = 1+\sqrt{2}.) Still, thanks for the answer, it really cleared things up a bit. $\endgroup$ – Jonas De Schouwer Apr 2 at 21:15

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