Find all possible values for the sum of the digits of a square.
Solution
Answer: any non-negative integer = 0, 1, 4 or 7 mod 9.
02 = 0, (±1)2 = 1, (±2)2 = 4, (±3)2 = 0, (±4)2 = 7 mod 9, so the condition is necessary.
We exhibit squares which give these values.
0 mod 9. Obviously 02 = 0. We have 92 = 81, 992 = 9801 and in general 9...92 = (10n - 1)2 = 102n - 2.10n + 1 = 9...980...01, with digit sum 9n.
1 mod 9. Obviously 12 = 1 with digit sum 1, and 82 = 64 with digit sum 10. We also have 982 = 9604, 9982 = 996004, and in general 9...982 = (10n - 2)2 = 102n - 4.10n + 4 = 9...960...04, with digit sum 9n+1.
4 mod 9 Obviously 22 = 4 with digit sum 4, and 72 = 49 with digit sum 13. Also 972 = 9409 with digit sum 22, 9972 = 994009 with digit sum 31, and in general 9...972 = (10n - 3)2 = 102n - 6.10n + 9 = 9...940...09, with digit sum 9n+4.
7 mod 9 Obviously 42 = 16, with digit sum 7. Also 952 = 9025, digit sum 16, 9952 = 990025 with digit sum 25, and in general 9...952 = (10n - 5)2 = 102n - 10n+1 + 25 = 9...90...025, with digit sum 9n-2.
© John Scholes
jscholes@kalva.demon.co.uk
22 Oct 2000