I Want Me More Unit Squares

Let the two sides of the grid (rectangle) be $a$ and $b$ , then the semiperimeter $s=a+b$ and the number of unit squares $n=ab$ . Without loss of generality, let us assume $b \ge a$ and their difference $b-a=d$ , so that $d \ge 0$ . Then, we have:

$\begin{aligned} s & = a+b = a+a+d = 2a + d\\ \implies a & = \frac {s-d}2 \\ \implies b & = \frac {s+d}2 \\ \implies n & = ab = \frac {s-d}2 \times \frac {s+d}2 = \frac {s^2-d^2}4 = \frac {s^2-(b-a)^2}4 \end{aligned}$

This means that for a particular $s$ the smaller the difference $d =b-a$ , the largest the $n$ . For $s=153$ , the smallest $d$ occurs when $b=57$ and $a=56$ and $d=57-56=1$ , then $U(153) = \dfrac {153^2-1}4 = \boxed{5852}$ .

In general, suppose the width of the grid is $a$ , the height of the grid is $b$ , and the semiperimeter of the grid is $s=a+b$ . Then the number of unit squares in this $a\times b$ grid is $ab$ .

We have 2 cases:

Case 1: $s$ is even.

When $s$ is even, then one of $a$ , $b$ must be greater than or equal to $\frac{s}{2}$ while the other must be smaller than or equal $\frac{s}{2}$ (they cannot be both greater than $\frac{s}{2}$ because $a+b$ would be greater than $s$ , they cannot be both smaller than $\frac{s}{2}$ either because $a+b$ would be smaller than $s$ ). Without loss of generality, let's suppose that $a=\frac{s}{2}+d$ , $b=\frac{s}{2}-d$ $(d<\frac{s}{2},\,d\in\mathbb{Z^+}\cup\{0\})$ . Then $\begin{aligned}ab&=\left(\frac{s}{2}+d\right)\left(\frac{s}{2}-d\right) \\ &=\frac{s^2}{4}-d^2 \\ &\leqslant\frac{s^2}{4}\end{aligned}$

From this we know that the maximum value of $ab$ for even $s$ is $\frac{s^2}{4}$ , equality can be achieved if and only if $d=0\implies a=b=\frac{s}{2}$ .

Case 2: $s$ is odd.

When $s$ is odd, then one of $a$ , $b$ must be greater than or equal to $\frac{s+1}{2}$ while the other must be smaller than or equal $\frac{s-1}{2}$ (for the same reason, they cannot be both greater than $\frac{s}{2}$ because $a+b$ would be greater than $s$ , and they cannot be both smaller than $\frac{s}{2}$ because $a+b$ would be smaller than $s$ ). Without loss of generality, let's suppose that $a=\frac{s+1}{2}+d$ , $b=\frac{s-1}{2}-d$ $(d<\frac{s-1}{2},\,d\in\mathbb{Z^+}\cup\{0\})$ . Then $\begin{aligned}ab&=\left(\frac{s+1}{2}+d\right)\left(\frac{s-1}{2}-d\right) \\ &=\frac{s^2-1}{4}-d^2 \\ &\leqslant\frac{s^2-1}{4}\end{aligned}$

From this we know that the maximum value of $ab$ for odd $s$ is $\frac{s^2-1}{4}$ , equality can be achieved if and only if $d=0\implies a=\frac{s+1}{2},\,b=\frac{s-1}{2}$ .

Therefore,

$U(s)=\begin{cases}\dfrac{s^2}{4},\quad\text{if }s\text{ is even} \\ \dfrac{s^2-1}{4},\quad\text{if }s\text{ is odd}\end{cases}$

Now, evaluating $U(153)$ , since $153$ is odd, we have $U(153)=\frac{153^2-1}{4}=5852$ .