Perimeter = Area

With the dimensions of the rectangle being $x,y \in \mathbb{Z^{+}}$ we require that $xy = 2x + 2y \Longrightarrow xy - 2x - 2y = 0 \Longrightarrow (x - 2)(y - 2) = 4$ .

The possible (unordered) factorings of $4$ to consider are $-4 \times -1, -2 \times -2, 2 \times 2$ and $4 \times 1$ , yielding respective solutions for $(x,y)$ of $(-2,1), (0,0), (4,4)$ and $(6,3)$ , only the latter $\boxed{2}$ representing suitable dimensions for a rectangle.

Since area and perimeter of rectangle has to be equal of dimensions $l$ and $b$ . And let us assume that $l \geq b$ then we can write as $l = b+ p$ where $p\geq 0$ . $\begin{aligned} & lb = 2(l+b) \\& b(b+p) = 2b + 2b +2p \\& b^2 +(p-4)b -2p =0 \\& b = \dfrac{-(p-4)\pm\sqrt{(p-4)^2 - 4(-2p)}}{2} \\& 2b = -(p-4)\pm{\color{#3D99F6}\sqrt{p^2 +16}}\cdots(1)\end{aligned}$ Shows that $b\in\mathbb Z^{+}$ iff $\sqrt{p^2+16}$ is perfect square which only possible for $p=\pm 3$ and $p=0$ respectively.

Replugging the possible values of $p\neq -3$ since $p\geq 0$ in the equation (1). The value turns out to be $b = 3$ and $b=4$ .

Therefore, the integer solution for rectangle for identical area and perimeter exist for $(l,b) = (6,3), (4,4)$ . So only 2 solutions exist.

I first set up the equation $xy=2(x+y)$ , where $x$ and $y$ must be integers. Then I realized one of $x$ or $y$ must be a multiple of $2$ . So, I chose $y=2k$ , where $k$ is an integer. $\Rightarrow 2kx=2(x+2k) \Rightarrow kx = x+2k \Rightarrow x=\frac{2k}{k-1}=2+\frac{2}{k-1}$ . The only values of $k$ that make $\frac{2}{k-1}$ an integer are $k=2,3$ . These values of $k$ then correspond to the two solutions $(x,y)=(4,4)=(3,6)$ .