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\begin{theorem}
It is possible to partition $\mathbb{R}^2$ into $25$ color classes such that none of them contains the vertices of a rectangle of area $1$.
\end{theorem}

\begin{proof}
Every rectangle of area $1$ is a special case of a parallelogram such that the product of lengths of its two consecutive sides equals $1$.
Thus, it is sufficient to give a coloring of $\mathbb{R}^2$ that uses $25$ colors and has a slightly stronger property: no color class will contain the vertices of a parallelogram such that the product of lengths of its two consecutive sides equals $1$.

Let $\mathbf{i}$ denote the complex imaginary unit.
Identify $\mathbb{R}^2$ with the complex plane $\mathbb{C}$.
Let us place a parallelogram $P=abcd$ in the complex plane, so that its vertices $a,b,c,d$ are respectively coordinatized by the complex numbers $z_a,z_b,z_c,z_d$.
Consider a complex quantity $I(P)$ defined as
$I(P) := z_a^2 - z_b^2 + z_c^2 - z_d^2$.
In this definition we specify the vertex $a$ to be the one with the smallest coordinate $z_a$ in the lexicographic ordering of $\mathbb{C}\equiv\mathbb{R}^2$.

There exist $u,v,z\in\mathbb{C}$ such that the vertices of $P$ have complex coordinates
\[ z_a=z, \quad z_b=z+u, \quad z_c=z+u+v, \quad z_d=z+v. \]
The quantity $I(P)$ now simplifies as
\[ I(P) = z^2 - (z+u)^2 + (z+u+v)^2 - (z+v)^2 = 2uv. \]
Consecutive side lengths of $P$ are $|u|$ and $|v|$, so we have
\[ |I(P)| = 2 \]
whenever their product equals $1$.
Therefore, it remains to find a coloring of $\mathbb{C}$ such that, if all vertices of $P$ are assigned the same color, then the complex number $I(P)$ does not lie on the circle
$\{w\in\mathbb{C} : |w|=2\}$.

For each pair $(j,k)\in\{0,1,2,3,4\}^2$ define a color class $C_{j,k}$ as
\[ C_{j,k} := \bigg\{ z\in\mathbb{C} : z^2 \in \frac{10}{3} \bigg( \mathbb{Z} + \mathbf{i} \mathbb{Z} + \frac{j + \mathbf{i} k}{5} + \big[0,\frac{1}{5}\big) + \mathbf{i} \big[0,\frac{1}{5}\big) \bigg) \bigg\}. \]
If the four vertices of $P=abcd$ belonged to the same color class, then, by the definition of $I(P)$, we would clearly have
\[ I(P) \in \frac{10}{3} \bigg( \mathbb{Z} + \mathbf{i} \mathbb{Z} + \big(-\frac{2}{5},\frac{2}{5}\big) + \mathbf{i} \big(-\frac{2}{5},\frac{2}{5}\big) \bigg). \]
The above set does not intersect the circle $\{w\in\mathbb{C} : |w|=2\}$.
Indeed, the central square lies fully inside the circle $\{w\in\mathbb{C} : |w|=2\}$ because of $4\sqrt{2}/3<2$, while all remaining open squares lie fully outside of that circle.
\end{proof}

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