Every two players play each other once. The outcome of each game is a win for one of the players. Player n wins a_{n} games and loses b_{n} games. Prove that ∑ a_{n}^{2} = ∑ b_{n}^{2}.

**Solution**

Suppose there are N players in total. Each player plays N-1 games, so b_{n} = N - 1 - a_{n}. Hence ∑ b_{n}^{2} = ∑ (N - 1)^{2} - 2(N - 1) ∑ a_{n} + ∑ a_{n}^{2} = N(N - 1)^{2} - 2(N - 1) ∑ a_{n} + ∑ a_{n}^{2}.

Each game is won by just one player, so ∑ a_{n} = no. of games = N(N - 1)/2. Hence ∑ b_{n}^{2} = N(N - 1)^{2} - 2(N - 1)N(N - 1)/2 + ∑a_{n}^{2} = ∑a_{n}^{2}.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002