. Find all positive integers $n$ such that the decimal representation of $n^{2}$ consists of odd digits only.
The only such numbers are $n=1$ and $n=3$.

If $n$ is even, then so is the last digit of $n^{2}$. If $n$ is odd and divisible by 5 , then $n=10 k+5$ for some integer $k \geq 0$ and the second-to-last digit of $n^{2}=(10 k+5)^{2}=100 k^{2}+100 k+25$ equals 2 .

Thus we may restrict ourselves to numbers of the form $n=10 k \pm m$, where $m \in\{1,3\}$. Then

$$
n^{2}=(10 k \pm m)^{2}=100 k^{2} \pm 20 k m+m^{2}=20 k(5 k \pm m)+m^{2}
$$

and since $m^{2} \in\{1,9\}$, the second-to-last digit of $n^{2}$ is even unless the number $20 k(5 k-m)$ is equal to zero. We therefore have $n^{2}=m^{2}$ so $n=1$ or $n=3$. These numbers indeed satisfy the required condition.