Method of ascent to prove that $x^2 − 3y^2 = 1$ has infinitely many solutions

Question

Use the method of ascent to prove there are infinitely many solutions to the Diophantine equation: $$x^2 − 3y^2 = 1$$ 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.

Answer 1

Consider the norm over $\mathbb{Z}[\sqrt{3}]$, $$ N(a+b\sqrt{3}) = (a+b\sqrt{3})(a-b\sqrt{3}) = a^2-3b^2. $$ The norm is multiplicative, hence if we found a unit (i.e. an element with unit norm) in $\mathbb{Z}[\sqrt{3}]$, from $N(z^n)=N(z)^n=1$ we get an infinite number of solutions for $a^2-3b^2 = 1$. For instance, the trivial solution $a=2,b=1$ gives that $(a,b)=(7,4)$ is also a solution, since:

$$ (2+\sqrt{3})^2 = 7+4\sqrt{3}. $$ As a consequence, if we consider the two sequences defined by: $$ a_1=2,\qquad a_2=7,\qquad a_{n+2}=4 a_{n+1}-a_{n}, $$ $$ b_1=1,\qquad b_2=4, \qquad b_{n+2}=4b_{n+1}-b_n, $$ we have that for every $n\geq 1$ $$ a_n^2-3b_n^2 = 1 $$ holds.