Squared Unit Squares!

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

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

Now, using the formula we have, we start by listing out some values to see if we can find any patterns:

$\begin{array}{|c|c|c|} \hline x & U(x) & U(U(x)) \\ \hline 2 & 1 & \text{undefinedillion}=1000^{\text{undefined}} \\ \hline 3 & 2 & 1 \\ \hline 4 & 4 & 4 \\ \hline 5 & 6 & 9 \\ \hline \color{orchid}6 & 9 & \color{orchid}20\\ \hline 7 & 12 & 36 \\ \hline 8 & 16 & 64 \\ \hline 9 & 20 & 100 \\ \hline \color{orchid}10 & 25 & \color{orchid}156 \\ \hline 11 & 30 & 225 \\ \hline 12 & 36 & 324 \\ \hline 13 & 42 & 441 \\ \hline \color{orchid}14 & 49 & \color{orchid}600 \\ \hline \end{array}$

Hey! We notice that if $x=4k+2$ $(k\in\mathbb Z^+)$ , then $U(U(x))$ breaks the perfect square pattern! Let's try to prove it!

For any integer $x\geqslant2$ , $x\equiv 0,\,\pm1,\,2\pmod4$ , that means $x$ can be expressed either as $4k$ , $4k\pm1$ or $4k+2$ .

If $x=4k$ , then since $4k$ is even, $U(4k)=\frac{(4k)^2}{4}=4k^2$ is even, thus $U(U(4k))=U(4k^2)=\frac{(4k^2)^2}{4}=(2k^2)^2$ is a perfect square.

If $x=4k\pm1$ , then since $4k\pm1$ is odd, $U(4k\pm1)=\frac{(4k\pm1)^2-1}{4}=4k^2\pm4k=4k(k\pm1)$ is even, thus $U(U(4k\pm1))=U(4k(k\pm1))=\frac{(4k(k\pm1))^2}{4}=(2k(k\pm1))^2$ is also a perfect square.

If $x=4k+2$ , then since $4k+2$ is even, $U(4k+2)=\frac{(4k+2)^2}{4}=(2k+1)^2$ is odd, thus $U(U(4k+2))=U((2k+1)^2)=\frac{(2k+1)^2-1}{4}=k^2+k=k(k+1)$ is not a perfect square. Whoopsy daisy!

If $x=\text{sandwich}$ , then... oh wait... nevermind.

Anyway, we've just showed that if $x=4k+2$ , in other words if $x\equiv2\pmod4$ , then $U(U(x))$ cannot be a perfect square! Woohoo! Progress!

Let's now take a look at the answer options, we notice some repeated numbers that we can use for our advantage. Since $1152^{1153}\equiv0\pmod4 \\ 1153^{1154}\equiv1^{1154}=1\pmod4 \\ 1154^{1155}\equiv0\pmod4 \\ 1155^{1156}\equiv(-1)^{1156}=1\pmod4 \\ 1156^{1157}\equiv0\pmod4$

The only option that gives a remainder of 2 when divided by 4 is $1153^{1154}+1155^{1156}$ , because $1153^{1154}+1155^{1156}\equiv1+1=2\pmod4$ .

By the way, the option "sixteen", otherwise famously known as $16$ , is obviously not the answer since $16\equiv0\pmod4$ .

$U(s)=\lfloor \frac{s}{2} \rfloor \times \lceil \frac{s}{2} \rceil$

here are facts

1- If $2 \mid s$ , then $U(s)=\frac{s^2}{4}$ is a perfect square.

2-If $s$ is odd, it would be of form $U(s)=n \times (n+1)$ , for some integer $n$ .

Now assume $s$ is odd. Then $U(s)=n \times (n+1)$ and either $n$ or $n+1$ is even ( $n \times (n+1)=2k$ ). Therefore, $U(n \times (n+1))=U(2k)=x^2$ , according to fact (1).

If $s$ is even, then $U(s)=\frac{s^2}{4}$ . If $\frac{s}{2}$ is not further divisible by $2$ , i.e. $4 \nmid s$ , then $U(U(s))=n \times (n+1)$ , for some $n$ (fact (2)). since two consecutive integers are coprimes, they cannot make perfect squares. On the other hand, if $4 \mid s$ , then we use rule (1), two times, to deduce that $U(U(s))$ is a square.

Based on the above argument, $U(U(s))$ is a perfect square, if and only if either $s$ is odd or $s$ divisible by $4$ .