Can someone help me to prove there are infinitely many solutions to the Diophantine equation: $$x^2 − 3y^2 = 1$$ using the method of ascent.
The Method of Ascent:
We can do this by showing how, given one solution $(u, v)$, we can compute another solution $(w, z)$ that is larger is some suitable sense. Then my proof will involve finding a pair of formulas, something like: $w = x + y$ and $z = x − y$. However I tried these formulas and they don't work. So I asked my teacher and she said that there is a pair of second degree formulas which do work; one of them has a cross term and one of them involves the number 3.
I asked this question earlier, but it was put on hold because it looked as a duplicate. However, I am asking to prove this using another method, the method of ascent which is quite different and has not yet been proven in such a way. Can someone please help me on this. I have been wanting an answer for a few days now. I know that the other page has the answer, but I don't care about the answer, I want to know how to do it.
The answers in the prior mentioned question imply that solutions may be composed by multiplying the quadratic integers $\,x_i+y_i\sqrt{3}\,$ associated with solutions $(x_i,y_i).\,$ In particular the powers of $\,2+\sqrt{3}\,$ generate an infinite increasing sequence of solutions (hence the ascent)
$$\begin{align} (2+\sqrt3)^2 &=\ \ \ \color{#c00}7\ +\ \color{#c00}4\sqrt3\ \ \Rightarrow\ \ \ \ 7^2-3\cdot 4^2\ \ = 1\\ (2+\sqrt3)^3 &=\ \color{#0a0}{26} + \color{#0a0}{15}\sqrt3\ \ \Rightarrow\ \ 26^2-3\cdot 15^2 = 1\\ (2+\sqrt3)^4 &=\ \color{#b0f}{97} + \color{#b0f}{56}\sqrt3\ \ \Rightarrow\ \ 97^2-3\cdot 56^2 = 1\\ &\ \ \vdots \end{align}$$
Or, using the composition law $\,(x,y) = \color{#c00}{(7,4)}\,\mapsto \,(2x+3y,2y+x) = \color{#0a0}{(26,15)},\ $ Then, again applying the composition law $(x,y) = \color{#0a0}{(26,15)}\,\mapsto \,(2x+3y,2y+x) = \color{#b0f}{(97,56)},\ \ldots$