2. Does there exist a natural number $n>3$ such that the decimal representation of the number $n^{2}$ consists only of odd digits? (20 points)
Solution: Suppose such a natural number $n>3$ exists. Then the number $n^{2}$ consists of at least two digits. Take the number formed by the last two digits of the number $n^{2}$, and call it $m$. According to the assumption, both digits of this number are odd. Therefore, the remainder when this number is divided by 4 is three. Then the remainder when the number $n^{2}$ is divided by 4 is also three, since $n^{2}-m=\overline{\ldots 00}: 4$. But the square of a number can only give remainders of 0 and 1 when divided by 4 - a contradiction. Therefore, such a number $n$ does not exist.

## Answer: does not exist