Sum of digits of perfect square

Question

Prove that the repeated sum of the digits of a perfect square can only be 1,4,7 or 9. Example- 169- Sum-16, Sum of digits of 16 = 7. Please try to explain without log, mods. I am a grade 10th student. Thank You.

Answer 1

Lemma 1: a natural number and the sum of its digits yield the same remainder when you divide them by $9$.

Lemma 2: if $n$ is a perfect square, the remainder of $n/9$ is $0,1,4$ or $7$.

Lemma 3: the repeated sum of the digits of a perfect square is $1,4,7$ or $9$.

It is not very hard to prove the first two lemmas, and the third should come easily with the first two.

Hint for Lemma 1: every natural number $m$ is $(a_k10^k-a_{k-1}10^{k-1}+\cdots+10a_1+a_0)$, where the $a_j$'s are its digits. Is $m-(a_k+a_{k-1}+\cdots+a_1+a_0)$ a multiple of $9$?