81. The square of an integer ends with four identical digits. Which ones?
81. A perfect square can only end in the digits 0, 1, 4, 9, 6, and 5. Furthermore, the square of any even number is clearly divisible by 4, while the square of any odd number gives a remainder of 1 when divided by 4 \(\left((2 k)^{2}=4 k^{2},(2 k+1)^{2}=4\left(k^{2}+k\right)+1\right)\); therefore, the square of no number can end in the digits 11, 99, 66, or 55 (numbers ending in two digits 11, 99, 66, or 55 give remainders of 3, 3, 2, and 3, respectively, when divided by 4). Finally, let's consider what remainders the square of an integer can give when divided by 16. Any integer can be represented in one of the following forms:

$$
8 k, 8 k \pm 1,8 k \pm 2,8 k \pm 3 \text { or } 8 k+4 \text {; }
$$

the squares of these numbers are of the form

$$
\begin{gathered}
16 \cdot\left(4 k^{2}\right), \quad 16\left(4 k^{2} \pm k\right)+1, \quad 16\left(4 k^{2} \pm 2 k\right)+4 \\
16\left(4 k^{2} \pm 3 k\right)+9 \quad \text { or } \quad 16\left(4 k^{2}+4 k+1\right) .
\end{gathered}
$$

Thus, we see that the square of an integer is either divisible by 16 or gives a remainder of 1, 4, or 9 when divided by 16. A number ending in the digits 4444 gives a remainder of 12 when divided by 16 and, therefore, cannot be a perfect square.

Therefore, if a perfect square ends in four identical digits, these digits must be four zeros (for example, \(100^{2}=10000\)).