Find all square numbers whose every digit is odd.
Solution. Among the single-digit square numbers, 1 and 9 are such. If a square number ends in an odd digit, then it can only be the square of an odd number. Let's examine the square of the odd numbers written in the form of a remainder when divided by ten! ( $k$ represents any positive integer)

$$
\begin{array}{ll}
(10 k+1)^{2}=100 k^{2}+20 k+1, & \text { in this case, the tens digit is even. } \\
(10 k+3)^{2}=100 k^{2}+60 k+9, & \text { in this case, the tens digit is even. } \\
(10 k+5)^{2}=100 k^{2}+100 k+25, & \text { in this case, the tens digit is even. } \\
(10 k+7)^{2}=100 k^{2}+140 k+49, & \text { in this case, the tens digit is even. } \\
(10 k+9)^{2}=100 k^{2}+180 k+81, & \text { in this case, the tens digit is even. }
\end{array}
$$

Thus, in the square of every odd number greater than 10, the digit in the tens place is even, so there are no more square numbers that meet the conditions.