1. Decrypt the equation

$$
\overline{a b c d}=(5 c+1)^{2}
$$
Solution. According to the condition of the problem, we have

$$
1000 a+110 b+10 c+d=25 c^{2}+10 c+1
$$

i.e.

$$
25\left(40 a+4 b-c^{2}\right)=1-d
$$

The left side is divisible by 25, so $25 \mid 1-d$, from which we get $d=1$. According to this, $40 a+4 b-c^{2}=0$, i.e. $4(10 a+b)=c^{2}$, from which we get $4 \mid c^{2}$. However, $a, b, c$ are digits and since $a \neq 0$, we immediately get that $c^{2}=64$ and $10 a+b=16$. Therefore, $c=8, b=6, a=1$ and the desired number is $\overline{a b c d}=1681$.