Problem - Sum of digits of $n$ and its square

Question

How can I solve this problem : $z(n)$ is the sum of the digits of $n$. So $z(15) = 1+5 = 6$

For all natural number $m$, there's an integer $n$ such that : $z(n)*m = z(n^2)$ ? You can find these integers here : https://oeis.org/A224792 Note : $n > 0$

Answer 1

Prove by induction: for any positive integer $m$ there is a number $n$ with all digits $0$ and $1$, $z(n)=m$ and $z(n^2) = m^2$.

For the induction step, note that if $n$ works for $m$, then $10^k + n$ works for $m+1$ if $k$ is sufficiently large.