Cryptogram AB x AB = ACD

Obviously $A$ can only be $1$ . Because if $A=2$ or $3$ , the hundreds digit of the product would be $4$ or $9$ respectively and if $A \ge 4$ , there would be a thousands unit. This means that $\overline{AB} \le \left \lfloor \sqrt{199} \right \rfloor = 14$ . Since $10^2 = 100$ , $11^2 = 121$ , $12^2 = 144$ , $13^2 = 169$ , and $14^2 = 196$ . Only $\boxed 2$ solutions $\overline{AB} = 13$ and $14$ satisfy the cryptogram.

For $\overline {AB}>14$ , there is no two digit number whose square is a three digit number with the hundred's place digit $A$ .

So, we have to check only five numbers : $10,11,12,13,14$

We see that only $\boxed {13}$ and $\boxed {14}$ are the numbers satisfying the given cryptogram.

$\begin{aligned} (10A + B) \times (10A + B) &= 100A + 10C + D \\ 100A^{2} + 20AB + B^{2} &= 100A + 10C + D \\ 10A^{2} + 2AB + B^{2} &= 10A + C + \dfrac{D}{10}\end{aligned}$

I can't complete it algebraically so by Trial and Error -

$13^2 = 169 \\ 14^{2} = 196 \\ \text{an we have our 2 solutions :)}$

I just write the answer directly

13² = 169

14² = 196