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Is it possible to determine when $$4a^2+4b^2+4a+4b+1 = ((2a+2b)+1)^2-8ab = (2a+1)^2+4(b^2+b)$$ is a perfect square, assuming $a,b \in \mathbb{Z}$ and $b>0$? I've tried writing it in a few different forms but I'm not sure if there's a strategy to approach this.

A.B
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    You can post your different forms as "context". – WhatsUp Jun 26 '22 at 01:42
  • The most promising idea I can think of is writing $(2a+1)^2+(2b+1)^2-1$. Then the problem reduces to finding solutions to $m^2+n^2=p^2+1$ with $m,n$ odd (and thus $p$ odd by mod $2$). – Connor Gordon Jun 26 '22 at 01:55
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    you have $x^2 + y^2 = 1 + z^2$ where $x,y$ and therefore $z$ are odd. There are infinitely many nontrivial answers. – Will Jagy Jun 26 '22 at 01:57
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    Does this answer your question? Solution of Diophantine equation - found using an Approach0 search. Note your result is odd, say it's $(2c + 1)^2$. Since your expression can also be written as $(2a + 1)^2 + (2b + 1)^2 - 1$, we get $(2a + 1)^2 + (2b + 1)^2 = (2c + 1)^2 + 1$. This is the proposed duplicate question with $a$ and $c$ switched around (note Will Jagy's comment above also indicates this). – John Omielan Jun 26 '22 at 01:59
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    one of my answers on this https://math.stackexchange.com/questions/2331734/infinitely-many-positive-integers-n-such-that-n21-mid-1-cdot-2-cdot-5-c/2331798#2331798 – Will Jagy Jun 26 '22 at 02:01
  • @A.B FYI, note that Integral solutions of hyperboloid $x^2+y^2-z^2=1$ is about a somewhat more general question (i.e., where the $3$ variables don't have to all be odd), with it having quite a few answers & different approaches. – John Omielan Jun 26 '22 at 02:03

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