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A certain rectangle has lengths $(2a-1)$ and $(2b-1)$, where $a$ and $b$ are integers. This implies that the lengths are odd integers.

The area of this rectangle is $(2a-1)(2b-1)$. A theorem claims that the area of this rectangle is equal to the difference bewtween two perfect squares, or $(c^2 -d^2)$ where $c$ and $d$ are integers.

How may I prove the theorem in the following form?

$$(2a-1)(2b-1)=(c^2-d^2)$$

or $$\forall a\in\mathbb{Z}, \forall b \in \mathbb{Z},\exists c \in \mathbb{Z},\exists d \in \mathbb{Z}, (2a-1)(2b-1)=(c^2-d^2)$$

I am aware that $c^2-d^2$ is equal to $(c-d)(c+d)$ and $(c-d)^2+2d(c-d)$, but I am not sure if this is at all helpful.

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Theorem, if two numbers $a$ and $b$ have the same parity then $ab$ is a difference of squares.

Proof: Notice that $a-b=2k$ for some $k$, now notice that $ab=((a+k)-k))((a+k)+k)=(a+k)^2-k^2$ as desired.

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    You are really looking at the squares of (a+b)/2 and (a-b)/2, but to have these be whole numbers you need that a and b have the same parity – 2017-02-21