Oddities within Perfection

Let's look the last two digits. A square number's last digit should be 0, 1, 4, 5, 6, 9, so the possible endings are: 11, 15, 19, 31, 35, 39, 51, 55, 59, 71, 75, 79, 91, 95, 99. A number can have 0, 1, 2 or 3 remainder when it is divided by four, so the square can have 0 or 1 remainder (mod 4). But the 11, 15, 19, 31, 35, 39, 51, 55, 59, 71, 75, 79, 91, 95, 99 numbers have 3 remainder by four. The only chance to have a square number with only odd digits, if the number don't have more than one digit. The one-digit square numbers are 1 and 9. So the answer is $\boxed{2}$ .

Besides 1 and 9, every perfect square has at least 1 even digit.

I will prove this by proving the following statement: Every perfect square's last two digits have at least 1 even digit (including the 0 in 01, 09, etc).

Now we only have to do the arithmetic mod 100. We can test every number between 0 and 99 and verify this result. This is because, for example, 43^2 has the same last two digits as 143^2, 243^2, etc. However, with some math, we can reduce the work a bit more.

Note that every 2-digit number can be written as 10a+b (ex. 63 = 10*6+3). We are looking for

$(10a+b)^2 = 100a +20ab+b^2 = 20ab + b^2 mod 100$

The last step comes from the fact that we only care about the last 2 digits, so the 100a doesn't matter to us. Also note that the 20ab will add an even quantity to the second digit. This will not change if any of the last two digits is odd, so we can ignore it. The only term that we need to be concerned with is the b^2, therefore, we only need to check every number between 0 and 9. We find that each has at least 1 even digit. Therefore, no perfect square with 2 or more digits has no even digit, and the only solutions are 1 and 9, so the answer is 2.