Given any positive integer k, show that we can find a1 < a2 < a3 < ... such that a1 = k and (a12 + a22 + ... + an2) is divisible by (a1 + a2 + ... + an) for all n.
Solution
Induction on n. Suppose a1 + a2 + ... + an = A divides a12 + a22 + ... + an2 = B. Then put an+1 = A2+B-A. We have a1 + a2 + ... + an+1 = A2+B, and an+12 = (A2+B)2 - 2A(A2+B) + A2, so a12 + a22 + ... + an+12 = (A2+B)2 - 2A(A2+B) + (A2+B), which is divisible by A2+B.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
7 February 2004
Last corrected/updatd 7 Feb 04