I have the equation $(4n+3)^2-48m^2=1$ that I changed into $X^2-48m^2=1$ so that it would be a Pell's equation, that has infinitely many solutions, that I found being $\left\{\begin{align} x_k = \frac{(7+4\sqrt{3} )^k + (7 - 4\sqrt{3})^k}{2}\\ y_k = \frac{(7+4\sqrt{3})^k - (7 - 4\sqrt{3})^k}{2\cdot4\sqrt{3}} \end{align}\right. , k\in\mathbb{N}\quad $ . How can I show that $(4n+3)^2-48m^2=1$ has an infinite numbers of solutions $\equiv3\pmod4$ ?
Thank you very much for help.
To summarize the discussion in the comments:
The algebraic numbers $7\pm 4\sqrt 3$ have minimal polynomial $$x^2-14x+1$$
It follows that the $x_k$ and the $y_k$ both satisfy the linear recursion $$a_n=14a_{n-1}-a_{n-2}$$
Now, $x_0=1, x_1=7$ and working $\pmod 4$ we see from that recursion that the terms alternate $\pm 1\pmod 4$. Thus there are infinitely many $x_n$ of each type.