Laws of divisibilities comes in pairs

Divisibility rules of $3$ and $9$ tell us that all of the numbers are divisible by $3$ , but not by $9$ , hence $\boxed{0}$ of them are square numbers.

Theorem

All perfect squares have digital roots of either $1$ , $4$ , $7$ , or $9$ . It must be noted, however, that the converse, (all numbers with digital roots of $1$ , $4$ , $7$ , or $9$ are perfect squares) is not true.

Proof

By the division algorithm, every integer $n$ is of the form $9k + r$ , where $0 \leq r < 9$ . So $n \equiv r \pmod 9$ and hence $n^2 \equiv r^2 \pmod 9$ .

Since $r \equiv r - 9 \pmod 9$ ,

$\displaystyle \begin{aligned} 0^2 & \equiv 0 \pmod{9} \\ \left(\pm 1\right)^2 & \equiv 1 \pmod{9} \\ \left(\pm 2\right)^2 & \equiv 4 \pmod{9} \\ \left(\pm 3\right)^2 & \equiv 0 \pmod{9} \\ \left(\pm 4\right) & \equiv 7 \pmod{9} \end{aligned}$

Thus, $n^2$ is congruent to $0$ , $1$ , $4$ , or $7$ , so its digital root is $1$ , $4$ , $7$ , or $9$ .

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Now suppose all possible 5-digit combinations of $12345$ were perfect squares. Their digital roots would all be $1 + 2 + 3 + 4 + 5 = 15 \rightarrow 1 + 5 = 6$ . Since the only digital roots of perfect squares are $1$ , $4$ , $7$ , or $9$ , we conclude by contradiction that $12345$ and all of its 5-digit combinations cannot be perfect squares.