Pell Equations: $a^2+4=5b^2$ This is a challenge problem in the Pell Equations chapter of my number theory book, but I'm not seeing the connection to Pell Equations. The Pell Equation with the coefficient $5$ is $5b^2+1=a^2$, but it doesn't look like the one I have.
Thanks if you can help me.

I know full well the problem was asked and answered 3 years ago.  Yet I wish to contribute.  Please critique and comment so I learn something.
$$\begin{align}
a^2+4&=5b^2 \\
a^2-5b^2&=-4 \quad \text{initial solution} \quad (a_1, b_1) =(1,1)\\
(a-b \sqrt5)(a+b \sqrt 5)&=-4\\
\text{since} \quad &(1-\sqrt 5)(1+\sqrt 5)=-4 \\
\text{then} \quad &(1-\sqrt 5)^2(1+\sqrt 5)^2=(-4)^2 \\
&(6-2 \sqrt 5)(6+2 \sqrt 5)=4^2 \quad \text{and}\\
(a-b \sqrt5)(6-2 \sqrt 5)(a+b \sqrt 5)(6+2 \sqrt 5)&=-4^3\\
(6a+10b)^2-5(2a+6b)^2&=-4^3 \\
\left(\frac{3a+5b}{2}\right)^2-5\left(\frac{a+3b}{2}\right)^2&=-4 \implies\\
&\begin{cases} 
a_{k+1}=\frac{1}{2}(3a_k+5b_k) \\ 
b_{k+1}= \frac{1}{2}(a_k+3b_k)\end{cases}\\
&\implies \\
&(1,1) \to (4,2) \to (11, 5) \to (29, 13) \to  \cdots
\end{align}$$
