Squares with All Odd Digits

Theorem
The only squares whose digits are all odd are $$1$$ and $$9$$.

Proof
If $$n$$ is even, then at least the last digit of $$n^2$$ is even.

So for $$n^2$$ to have its digits all odd, $$n$$ itself must be odd.

We can see immediately that $$1 = 1^2$$ and $$9 = 3^2$$ fit the criterion.

Of the other 1-digit odd integers, we have $$5^2 = 25, 7^2 = 49, 9^2 = 81$$, all of which have an even digit.

Now, let $$n > 10$$ be an odd integer. There are five cases to consider:


 * $$n = 10 p + 1$$: we have $$\left({10p + 1}\right)^2 = 100 p^2 + 20 p + 1 = 10 \left({10 p^2 + 2 p}\right) + 1$$.


 * $$n = 10 p + 3$$: we have $$\left({10p + 3}\right)^2 = 100 p^2 + 60 p + 9 = 10 \left({10 p^2 + 6 p}\right) + 9$$.


 * $$n = 10 p + 5$$: we have $$\left({10p + 5}\right)^2 = 100 p^2 + 100 p + 25 = 10 \left({10 p^2 + 10 p + 2}\right)+ 5$$.


 * $$n = 10 p + 7$$: we have $$\left({10p + 7}\right)^2 = 100 p^2 + 140 p + 49 = 10 \left({10 p^2 + 14 p + 4}\right)+ 9$$.


 * $$n = 10 p + 9$$: we have $$\left({10p + 9}\right)^2 = 100 p^2 + 180 p + 81 = 10 \left({10 p^2 + 18 p + 8}\right)+ 1$$.

It is clear that in all cases the 10's digit is even.

So the square of every odd integer greater than $$3$$ always has at least one even digit.

Hence the result.