Equation of a rectangle
Based on Raskolnikov's answer here, one can build an implicit Cartesian equation for a $2p \times 2q$ rectangle:
$$\left(\frac{x}{p}\right)^2+\left(\frac{y}{q}\right)^2=\sec\left(\arctan\left(\frac{x}{p},\frac{y}{q}\right)-\frac{\pi}{2}\left\lfloor\frac2{\pi}\arctan\left(\frac{x}{p},\frac{y}{q}\right)+\frac12\right\rfloor\right)^2$$
Another one is based on modifying the implicit equation of a Lamé curve:
$$\left|\frac{x}{p}+\frac{y}{q}\right|+\left|\frac{x}{p}-\frac{y}{q}\right|=2$$
For purposes of plotting with a computer, the implicit equation isn't terribly convenient to handle, so I'll throw in a set of parametric Cartesian equations for free, based on the parametric equations of the Lamé curve:
$$\begin{align*}x&=p\left(|\cos\,t|\cos\,t+|\sin\,t|\sin\,t\right)\\y&=q\left(|\cos\,t|\cos\,t-|\sin\,t|\sin\,t\right)\end{align*}$$
Here's another one, based on a special case of the parametric equations given in this answer:
$$\begin{align*}x&=p\left(\cos\left(\frac{\pi}{2}\lfloor u\rfloor\right)-(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\\y&=q\left(\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)+(2u-2\lfloor u\rfloor-1)\cos \left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\end{align*}$$
...and another one:
$$\begin{align*}x&=p\max\left(-1,\min\left(\frac4{\pi}\arcsin\left(\sin\left(\frac{\pi u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\\y&=q\max\left(-1,\min\left(-\frac4{\pi}\arcsin\left(\cos\left(\frac{\pi u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\end{align*}$$
...and I suppose I should stop here. ;)
This is an equation for a rectangle which has corners at $(a,b)$ and $(c,d)$
$$(x-a)(x-c)(y-b)(y-d)=0$$
but it extends a little beyond the corners, so instead
$$\sqrt{(a-x)(x-c)}\sqrt{(b-y)(y-d)}=0$$
which would throw an error for square roots of negative numbers
Try plotting $x^n + y^n = p^n$ where $p$ is the side length and $n$ is an even number. The larger $n$ is, the sharper the sides are.