Positive integer solutions to $5a^2 + 8a + 4 = 4b^2$ I'm convinced $5a^2 + 8a + 4 = 4b^2$ can be somehow turned into Pell's equation. 
My first steps: Rewrite as $5(a + 4/5)^2 + 4/5 = 4b^2$. Rewrite $B = 2b, A = 5a + 4$ to get $A^2 + 4 = 5B^2$. I'm almost there, I just need to know how to solve this Pell-like equation.

user236182's answer: $a$ must be even, $a = 2k$. Rewrite as $5(4k^2) + 16k +4 = 4b^2$, then multiply by $5/4$ to get $25k^2 + 20k + 5 = 5 b^2 \iff (5k + 2)^2 - 5b^2 = -1$. This is Pell's equation in the form $x^2 - 5y^2 = -1$. This is sufficient for all positive integer solutions. 
