Let's write numbers in place of the letters in the equation

$$
\overline{a b c}^{2}=\overline{a d e f f}
$$

such that the equation is true. (The overline indicates that the letters represent digits in a decimal number system. Identical letters represent identical digits, and identical digits are represented by identical letters.)
I. Solution: 1. Since $\overline{a b c}$ squared is a five-digit number, $\overline{a b c}<\sqrt{100000}<317$.

2. Since the first digit of the five-digit square is also $a$, then $a=1$.
3. During squaring, $2 a \cdot b=2 b<10$ (otherwise $\overline{1 b c}^{2}$ would start with a number greater than 1); this implies that $b<5$.
4. Squaring $bc$, the last two digits are equal $(f=f)$, and since the tens place of $2 b \cdot c$ is always even, $c^{2}$ must have an even last digit. This is only satisfied by $c=0,2,8$.
5. However, $c \neq 0$, because $c \neq f$. If $c=2$, then $f=4$, and since the last digit of $2 b \cdot c$ is $f=4$, then $b=1$ (since $b<5$), but $b \neq 1=a$, so $c \neq 2$.

Therefore, $c=8$, and $f=4$.
6. $b<5$; but $b \neq 4=f, b \neq 1=a, b \neq 0$, because in this case $\overline{a b c}^{2}$ cannot end in 44.

So $b$ can be 2 or 3.

Since $128^{2}=16384$ does not satisfy the condition, the only solution is:

$$
138^{2}=19044
$$

Magdolna Tisza (Debrecen, Kossuth lg. III. o. t.)

II. Solution: Since $\overline{a b c}$ squared is only a five-digit number and also starts with $a$, then $a=1$, and thus $\overline{a b c}<\sqrt{20000}<142$. Therefore, $\overline{b c}<42$.

A square number can only end in two identical digits if the number itself ends in (considering $\overline{b c}<42$): $00,12,38$.

But 00 is not possible, because $c \neq f$, 12 is not possible, because $b \neq 1=a$, and thus only 38 satisfies the conditions.

Indeed,

$$
\overline{a b c^{2}}=138^{2}=\overline{a d e f f}=19044
$$

Piroska Rétey (Debrecen, Svetits lg. II. o. t.)