<s>[INST] Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:
Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)
I know one of solutions is $5776=(76)^2$.
 [/INST] $ABCD=(CD)^2$ can be translated as $100n + x = x^2$ which implies 
$x^2 - x \equiv 0 \pmod{100}$.
$$ \text{$x^2-x \equiv 0 \pmod{100} \iff x(x-1) \equiv 0 \pmod{4}$ 
    and $x(x-1) \equiv 0 \pmod{25}$}$$
$$x^2-x \equiv 0 \pmod{4} \iff x \in \{\bar 0_{4}, \bar1_{4} \}$$
$$x^2-x \equiv 0 \pmod{25} \iff x \in \{\bar 0_{25}, \bar1_{25} \}$$
$$x\equiv a \pmod{4} \ \wedge \ x \equiv b \pmod{25} \iff x \equiv 25a - 24b \pmod{100}$$
\begin{align}
   x \equiv 0 \pmod{4} \ \wedge \ x \equiv 0 \pmod{25} \iff x \equiv 0 \pmod{100} \\
   x \equiv 0 \pmod{4} \ \wedge \ x \equiv 1 \pmod{25} \iff x \equiv 76 \pmod{100} \\
   x \equiv 1 \pmod{4} \ \wedge \ x \equiv 0 \pmod{25} \iff x \equiv 25 \pmod{100} \\
   x \equiv 1 \pmod{4} \ \wedge \ x \equiv 1 \pmod{25} \iff x \equiv 1 \pmod{100} \\
\end{align}
So, we get


*

*00^2 = 0000

*01^2 = 0001

*25^2 = 0625

*76^2 = 5776

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