Digital Root of Square

Theorem

Let $n^2$ be a square number.

Then the digital root of $n^2$ is $1$, $4$, $7$ or $9$.

Proof

Let $\map d n$ denote the digital root base $10$ of $n$.

From Digital Root is Congruent to Number Modulo Base minus 1, $\map d n \equiv n \pmod 9$.

So, let $n = 9 k + m$ where $1 \le m \le 9$.

Thus:

\(\ds n^2\)

\(=\)

\(\ds \paren {9 k + m}^2\)

\(\ds \)

\(=\)

\(\ds 81 k^2 + 18 k m + m^2\)

\(\ds \)

\(=\)

\(\ds 9 \paren {9 k^2 + 2 k m} + m^2\)

\(\ds \)

\(\equiv\)

\(\ds m^2\)

\(\ds \pmod 9\)

We enumerate the squares of the digits:

\(\ds 1^2\)

\(=\)

\(\ds 1\)

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 9\)

\(\ds 2^2\)

\(=\)

\(\ds 4\)

\(\ds \)

\(\equiv\)

\(\ds 4\)

\(\ds \pmod 9\)

\(\ds 3^2\)

\(=\)

\(\ds 9\)

\(\ds \)

\(\equiv\)

\(\ds 9\)

\(\ds \pmod 9\)

\(\ds 4^2\)

\(=\)

\(\ds 16\)

\(\ds \)

\(\equiv\)

\(\ds 7\)

\(\ds \pmod 9\)

\(\ds 5^2\)

\(=\)

\(\ds 25\)

\(\ds \)

\(\equiv\)

\(\ds 7\)

\(\ds \pmod 9\)

\(\ds 6^2\)

\(=\)

\(\ds 36\)

\(\ds \)

\(\equiv\)

\(\ds 9\)

\(\ds \pmod 9\)

\(\ds 7^2\)

\(=\)

\(\ds 49\)

\(\ds \)

\(\equiv\)

\(\ds 4\)

\(\ds \pmod 9\)

\(\ds 8^2\)

\(=\)

\(\ds 64\)

\(\ds \)

\(\equiv\)

\(\ds 1\)

\(\ds \pmod 9\)

\(\ds 9^2\)

\(=\)

\(\ds 81\)

\(\ds \)

\(\equiv\)

\(\ds 9\)

\(\ds \pmod 9\)

Hence the result.

$\blacksquare$