Unit Squares Galore

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 $n$ . Since for even $s$ (see Note for odd $s$ ) the smallest $d=0$ , when $b=a$ , $n_{max} = U(s) = \dfrac {s^2}4$ . The largest $d$ occurs when $a$ is smallest, that is $a=1$ . Since $s=a+b \implies b = s-1$ and $d = b-a = s-2$ when $n$ is minimum. Therefore, $n_{min} = u(s) = \dfrac {s^2 - (s-2)^2}4 = s-1$ .

Therefore, we have:

$\begin{aligned} U(x) - u(x) & = 49 \\ \frac {x^2}4 - x + 1 & = 49 \\ x^2 - 4x - 192 & = 0 \\ (x-16)(x+12) & = 0 \\ \implies x & = \boxed{16} & \small \color{#3D99F6} \text{since }x > 0\end{aligned}$

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Note: For odd $s$ , the smallest $d=1$ , when $b=a+1$ . The largest $d$ occurs when $a$ is smallest, that is $a=1$ . For example if $x=5$ , for $U(5)$ , we have $b=3$ and $a=2$ some that $d=1$ , therefore $U(5)=6$ . And for $u(5)$ , $a=1$ and $b=4$ , therefore $u(5) = 4$ .

First of all, I claim that $U(s)=\begin{cases}\dfrac{s^2}{4},\quad\text{if }s\text{ is even} \\ \dfrac{s^2-1}{4},\quad\text{if }s\text{ is odd}\end{cases}$ $u(s)=s-1$

Here is a sketch of the proof, you can find the full details of this proof in my solution to I Want Me More Unit Squares and I Want Me Less Unit Squares , if you've already seen it, you can safely skip this part. :D

Proof:

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 $ab$ .

If $s$ is even , either $a\geqslant\frac{s}{2}$ , $b\leqslant\frac{s}{2}$ or $a\leqslant\frac{s}{2}$ , $b\geqslant\frac{s}{2}$ . 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 $ab=\left(\frac{s}{2}+d\right)\left(\frac{s}{2}-d\right)=\frac{s^2}{4}-d^2\leqslant\frac{s^2}{4}$ .

If $s$ is odd , either $a\geqslant\frac{s+1}{2}$ , $b\leqslant\frac{s-1}{2}$ or $a\leqslant\frac{s-1}{2}$ , $b\geqslant\frac{s+1}{2}$ . 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 $ab=\left(\frac{s+1}{2}+d\right)\left(\frac{s-1}{2}-d\right)=\frac{s^2-1}{4}-d^2\leqslant\frac{s^2-1}{4}$ .

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

On the other hand, we have $ab=\frac{(a+b)^2-(a-b)^2}{4}=\frac{s^2-(a-b)^2}{4}$ , to minimize $ab$ , we need to minimize $(a-b)^2$ , this can be achieved with $a=s-1$ , $b=1$ , making $ab=\frac{s-(s-2)^2}{4}=s-1$ .

Therefore, $u(s)=s-1$ . $_\square$

Now, from $U(x)-u(x)=49$ , we have $U(x)-(x-1)=49\implies U(x)-x=48$

Since $U(x)$ and $x$ are both integers and $48$ is even, this implies that $U(x)$ , $x$ are either both even or both odd.

Let's assume that $U(x)$ and $x$ are both odd, let $x=2k+1$ where $k$ is a positive integer, then $U(2k+1)=\frac{(2k+1)^2-1}{4}=k^2+k=k(k+1)$

Because $k(k+1)$ is always even, this tells us that $U(x)$ is even, but we just assumed that it was odd, a contradiction! Oh noes!

Therefore, $U(x)$ and $x$ must be both even, hence from $U(x)-x=48$ , we have the following $\frac{x^2}{4}-x=48\implies x^2-4x-192=0\implies (x+12)(x-16)=0$

Noting that $x>0$ , solving this, we get $x=16$ .