What are the possible solutions for the diophantine equation $4x^2-3y^2=1$ and is there a general formula?

Question

Assuming that $a = x^2$ and $b = y^2$, i converted this equation to a linear diophantine equation for sake of convenience:

$$4a - 3b = 1$$

where after calculating a particular solution (like $(1, 1)$) by extended euclidean algorithm, the $n$th pair of solutions can be calculated as

$$a_n = 1 - 3n$$ $$b_n = 1 - 4n$$

where $n \le 0$ since $a_n, b_n$ are equal to squared numbers and so they have to be equal or greater than zero. So let the:

$$ 0 <= m = -n$$ $$a_m = 3m + 1$$ $$b_m = 4m + 1$$

Then for a pair of $(a_m, b_m)$ where $a_m, b_m$ are perfect squares, the pair $(\sqrt{a_m}, \sqrt{b_m})$ will be a solution for our diophantus equation. Written more generally:

$$(\sqrt{a_m}, \sqrt{b_m}) = (\sqrt{3m + 1}, \sqrt{4m + 1})$$ $$\exists m\in \mathbb Z$$

Of course only when both $3m + 1$ and $4m + 1$ are perfect squares. And this is the point i came at. I don't know what to do after that. Any help would be appreciated.

By the way i wrote a Python script to compute solutions by the method i mentioned above. Here are few pair i succeed to calculate:

$${(1.0, 1.0), (13.0, 15.0), (181.0, 209.0), (2521.0, 2911.0), ...}$$

I can't see any connection among the pairs... How i can continue?

Answer 1

or, we have solution pairs $(x_n, y_n)$ with $$ x_{n+2} = 14 x_{n+1} - x_n \; , \; $$ $$ y_{n+2} = 14 y_{n+1} - y_n \; , \; $$

The sequence of $x$ values begins $$ 1, 13, 181, 2521, 35113, 489061, 6811741, ... $$ The $y$ are $$ 1, 15, 209, 2911, 40545, 564719, 7865521, ... $$

Your form discriminant is $\Delta =0^2 - 4 \cdot 4 \cdot (-3) = 48.$ The defining problem for all oriented "automorphism" matrices is $$ \tau^2 - \Delta \sigma^2 = 4. $$ Then, with coefficients $\langle A,B,C \rangle,$ an automorphism matrix is $$ \left( \begin{array}{cc} \frac{\tau - B \sigma }{2} & -C \sigma \\ A \sigma & \frac{\tau + B \sigma }{2} \end{array} \right) $$ With the minimum $\tau = 14, \sigma = 2$ we get $$ \left( \begin{array}{cc} 7 & 6 \\ 8 & 7 \end{array} \right) $$ Indeed, you can check that if $$ u = 7 x + 6 y, \; \; v = 8x+7y, $$ then $$ 4 u^2 - 3 v^2 = 4 x^2 - 3 y^2 $$ The relation with the $14$ is just using Cayley-Hamilton on the matrix, which has trace $14$ and determinant $1.$

Answer 2

As noted, this is the same as $x'-3y^2=1$, known as Pell's equation ($x'=2x$). To find the first nontrivial solution (aka $1,0)$, we take $x'=2,y=1$. Then all solutions are of the form

$$x'=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}$$

$$y=\frac{(2+\sqrt{3})^n-(2-\sqrt{3})^n}{2\sqrt{3}}$$

And $x'$ is even. This occurs precisely when $n$ is odd, as the only terms that are left over in the binomial expansion for the numerator $\mod 4$ are $2n\sqrt{3}^{n-1}$ for $n$ odd, $2\sqrt{3}^n$ for $n$ even.

Thus , we take $x'=2,y=1$. Then all solutions are of the form

$$(x,y)=\left(\frac{(2+\sqrt{3})^{2n-1}+(2-\sqrt{3})^{2n-1}}{2},\frac{(2+\sqrt{3})^{2n-1}-(2-\sqrt{3})^{2n-1}}{2\sqrt{3}}\right)$$

is the final solution set as $n$ ranges over $\mathbb{N}$