Proof that the equation $x^2 - 3y^2 = 1$ has infinite solutions for $x$ and $y$ being integers

Based on a similar result I know for $x^2 -2y^2 = 1$, I would bet solid gold it falls out of abstract algebra in the study of the ring $\mathbb{Z}[\sqrt{3}] = \{ a + \sqrt{3}b \ | \ a, b \in \mathbb{Z} \}$.

We define a function $N: \mathbb{Z}[\sqrt{3}] \to \mathbb{Z}$ modeled on the complex norm, which multiplies a number by its complex conjugate.

$N(x + \sqrt{3}y) := (x +\sqrt{3}y)(x - \sqrt{3}y) = x^2 -3y^2$

Now, it just so happens that this shares a nice property of the complex norm.

For any two elements $z, w \in \mathbb{Z}[\sqrt{3}]$ we have $N(zw) = N(z)N(w)$. This isn't difficult to prove, just do the computation.

So, say we have some $x + \sqrt{3}y = z \in \mathbb{Z}[\sqrt{3}]$ with $N(z) = x^2 -3y^2 = 1$. It's easy to check that $(2, 1)$ is also a solution, and hence $N(2 + \sqrt{3}) = 1$.

Now we have $1 = N(z)N(2 + \sqrt{3}) = N((x + \sqrt{3}y)(2 + \sqrt{3})) = N((2x+3y) + \sqrt{3}(x + 2y))$.

But $N((2x+3y) + \sqrt{3}(x + 2y)) = 1$ means that $(2x+3y, x + 2y)$ is a solution.

If your initial values for x and y are both positive, then the solution they generate must be larger, so each new solution must be distinct.

I see a couple issues that may prove to be stumbling blocks. First you might have had problems verifying the solution $\,(\bar x,\bar y) = (2x+3y,x+\color{#c00}3y)\,$ because it is incorrect: the $\color{#c00}3$ should be $2.\,$ Now one can may verify that $\,(\bar x,\bar y)\,$ is also a solution using no more than simple integer arithmetic

$$\begin{eqnarray} \bar x^2 &=&\quad\ (2x+3y)^2 &=&\,\ \ \ 4x^2\,+\ 9y^2 +\, 12xy\\ -3\,\bar y^2&=&-3(x^2+2y)^2 &=&\, -3x^2-12y^2-12xy \\ \hline \bar x ^2 -\, 3\bar y^2&& &=& \ \ \ \ x^2\ -\ 3y^2 =\, 1 \end{eqnarray}\qquad\qquad $$

Though some explanations of the genesis of this composition law on the solution space may utilize ideas that you have not yet learned (such as the multiplicativity of the norm map on quadratic felds), the above direct proof that $\,(\bar x, \bar y)\,$ is a solution does not require such methods. Rather, it is a simple calculation using only integer arithmetic.

Said composition law is a special case of the fact that one may compose (multiply) any two solutions using the Brahmagupta–Fibonacci identity below. Above is the special case $\,a,b = 2,1.$

$$\begin{eqnarray} (a^2-3b^2)(x^2-3y^2) &\,=\,& (ax+3by)^2- 3(ay+bx)^2\\ (a+b\sqrt{3})(x+y\sqrt{3})\, &\,=\,& \ \,ax+3by\ \ +\,\ \ (ay+bx)\sqrt{3} \end{eqnarray}\qquad$$

Again, though the composition law may be intuitively derived by taking the norm of the quadratic integers listed below it, you can verify the integer identity above it using only integer arithmetic - independent of "irrational" numbers, quadratic fields, norms, etc.