After casting away nines, the result is 5, was the original integer a square?

Let the perfect square be $a^2$ or $a \times a$ . Adding up all the digits of $a$ (also called casting out nines since the result is the remainder after dividing by 9) gives us the “digital sum”. Let the digital sum of $a$ be $n$ . Then the wiki page says that the product of the digital sums ( $n \times n$ ) will be the digital sum of the original square, or 5. In short, $n^2=5$ But then $n$ won’t be a digit, which is impossible. Therefore a perfect square cannot have a digital sum equal to 5.

Consider the result of casting away nines from the squares of the integers from 0 to 9: $\left( \begin{array}{rrr} 0 & 0 & 0 \\ 1 & 1 & 1 \\ 2 & 4 & 4 \\ 3 & 9 & 9 \\ 4 & 16 & 7 \\ 5 & 25 & 7 \\ 6 & 36 & 9 \\ 7 & 49 & 4 \\ 8 & 64 & 1 \\ 9 & 81 & 9 \\ \end{array} \right)$

There is no 5 in the right hand column. Therefore, the original positive integer was not a square. The answer is No .