Solving Pell's equation $x^2-5y^2=\pm4$ using elementary methods. 
Solve Pell's equation $x^2-5y^2=\pm4$.

This equation arises when I tried to prove that the units of $\mathbb{Z}[\varphi]$, where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio, are of the form $\pm\varphi^n$. I found that $x+\varphi y$ is a unit iff $x^2+xy-y^2=\pm1$, i.e. $(2x+y)^2-5y^2=\pm4$. Yet I am unable to solve this equation. I saw here a solution using algebraic number theory, but I am interested in how to solve this equation using elementary methods, without using results from algebraic number theory. Thanks in advance!

For $x^2-5y^2=-4$ we have a base solution at $(x,y) = (1,1)$ and the unit solution for $x^2-5y^2=\pm1$ is $(x,y) = (2,1)$. So you can generate infinitely many solutions with:
\begin{equation*} x^2-5y^2 = (1-\sqrt{5})(1+\sqrt{5})(2-\sqrt{5})^n(2+\sqrt{5})^n = (-4)(-1)^n \end{equation*}
With $n=1$ that gives $(1+\sqrt{5})(2-\sqrt{5}) = -3 +\sqrt{5}, (1-\sqrt{5})(2+\sqrt{5}) = -3 -\sqrt{5}$ and we thus have $3^2-5(1^2) = 4$. We also have another base solution for $x^2-5y^2=4$ with $(x,y) = (\pm2, 0)$ which generates another family of solutions:
\begin{equation*} x^2-5y^2 = 4(2-\sqrt{5})^n(2+\sqrt{5})^n = (4)(-1)^n \end{equation*}
