For each fixed positive integer $N\in\mathbb{N}$, let's define two sets
\begin{align} A_N:=&\{(a,b)\in\mathbb{N}^2: N=a(2b-1)+(2a-1)(b-1)\}, \\ B_N:=&\{(c,d)\in\mathbb{N}^2: N=c(2d-1)+(d-1)(d-1)\}. \end{align}
QUESTION. For each $N$, is it true that $\#A_N=\#B_N$?
The answer is "yes" if by $\mathbb{N}$ we understand the set of positive integers.
Indeed, following the comment of Ekene E., \begin{align} A_N=&\{(a,b)\in\mathbb{N}^2: 4N-1=(4a-1)(4b-3)\}, \\ B_N=&\{(c,d)\in\mathbb{N}^2: 4N-1=(2d-1)(4c+2d-3)\}. \end{align} Now for any decomposition $4N-1=uv$ with $u,v\in\mathbb{N}$, one of $u$ and $v$ is congruent to $1\bmod{4}$, while the other is congruent to $3\bmod{4}$. So in exactly half of these decompositions, namely when $u\equiv 3\pmod{4}$ and $v\equiv 1\pmod{4}$, we can write $u=4a-1$ and $v=4b-3$ with $a,b\in\mathbb{N}$. Hence $\#A_N$ is half the number of divisors of $4N-1$. Similarly, for any decomposition $4N-1=uv$ with $u,v\in\mathbb{N}$, both $u$ and $v$ are odd, and $u-v$ is an integer congruent to $2\bmod{4}$. So in exactly half of these decompositions, namely when $u<v$, we can write $u=2d-1$ and $v=4c+2d-3$ with $c,d\in\mathbb{N}$. Hence $\#B_N$ is also half the number of divisors of $4N-1$.
