We can rewrite original equation as $(2a-b)^2- 5b^2 = -4$. We can assume that $b$ is even integer, it means that $b = 2b_1$ for some positive integer $b_1$. As a result we obtain $(a-b_1)^2-5b_{1}^2 = - 1$, which is negative Pell's equation with minimal solution $(2;1)$, which gives us that original equation has infinitely many solutions.

Fibonacci numbers??

soryn wrote: Fibonacci numbers?? Thank you for the idea. It might be the actual inspiration for this problem. We use the fact $F_{n-1}F_{n+1}-F_{n}^2=(-1)^n$ (easy induction) $(*)$ Rewrite $a(a-b)-b^2=-1$ Now we can choose $a=F_{2k+2}, b=F_{2k+1}$ for each $k \in \mathbb{N}$ We get $F_{2k+2}F_{2k}-F_{2k+1}^2=-1$, which is true from $(*)$

Kimchiks926 wrote: Prove that equation $a^2 - b^2=ab - 1$ has infinitely many solutions, if $a,b$ are positive integers Trivial by Vieta Jumping. Edit : Sorry @below , due to slip of mind , this equation seemed symmetric at the first glance.

@bumblebee60 - How do you employ vieta jumping... Agree initial seed solution is (1,1) But equation does not seem symmetry and how can we do root flipping?