Find all possible values for the sum of the digits of a square.
Answer: any non-negative integer $=0,1,4$ or $7 \bmod 9$.

$0^{2}=0, (\pm 1)^{2}=1, (\pm 2)^{2}=4, (\pm 3)^{2}=0, (\pm 4)^{2}=7 \bmod 9$, so the condition is necessary.

We exhibit squares which give these values.

$0 \bmod 9$. Obviously $0^{2}=0$. We have $9^{2}=81, 99^{2}=9801$ and in general $9 \ldots 9^{2}=(10^{\mathrm{n}}-1)^{2}=$ $10^{2 \mathrm{n}}-2 \cdot 10^{\mathrm{n}}+1=9 \ldots 980 \ldots 01$, with digit sum $9 n$.

1 mod 9. Obviously $1^{2}=1$ with digit sum 1, and $8^{2}=64$ with digit sum 10. We also have $98^{2}$ $=9604, 998^{2}=996004$, and in general $9 \ldots 98^{2}=(10^{\mathrm{n}}-2)^{2}=10^{2 \mathrm{n}}-4 \cdot 10^{\mathrm{n}}+4=9 \ldots 960 \ldots 04$, with digit sum $9 n+1$.

$4 \bmod 9$ Obviously $2^{2}=4$ with digit sum 4, and $7^{2}=49$ with digit sum 13. Also $97^{2}=9409$ with digit sum $22, 997^{2}=994009$ with digit sum 31, and in general $9 \ldots 97^{2}=(10^{\mathrm{n}}-3)^{2}=10^{2 \mathrm{n}}-$ $6 \cdot 10^{\mathrm{n}}+9=9 \ldots 940 \ldots 09$, with digit sum $9 n+4$.

$7 \bmod 9$ Obviously $4^{2}=16$, with digit sum 7. Also $95^{2}=9025$, digit sum $16, 995^{2}=990025$ with digit sum 25, and in general $9 \ldots 95^{2}=(10^{\mathrm{n}}-5)^{2}=10^{2 \mathrm{n}}-10^{\mathrm{n}+1}+25=9 \ldots 90 \ldots 025$, with digit sum $9 n-2$.