The Tournament Count

The problem

In a chess club every player plays every other player exactly once, and no game is drawn: each game has one winner and one loser. When the tournament is over, each player counts their wins, $w$, and their losses, $l$.

Prove that, whatever the results, the sum of the squares of all the win counts equals the sum of the squares of all the loss counts:

$w_1^2 + w_2^2 + \dots + w_n^2 = l_1^2 + l_2^2 + \dots + l_n^2.$

Before you begin

Run a tournament of four players on paper: draw the results grid and fill in the six games any way you like. Compute both sums. Then change one result and compute again. Both sums move, but they stay equal, and that should tell you the reason is structural, not arithmetical.

1. How many games are played altogether? Every game hands out exactly one win and one loss. What does that tell you about the sum of all the $w$'s compared with the sum of all the $l$'s?
2. Take any one player. What is $w + l$ for that player? Is it the same number for everybody?
3. $w^2 - l^2$ factorises. Put the two facts above into the factorisation.

Extension, for after the proof. Call three players a transitive triple if one of them beat both of the others. Count the transitive triples in two ways, once by the player who did the beating and once by the player who was beaten twice, and you have a second proof. As a bonus, express the number of cyclic triples ($A$ beats $B$, $B$ beats $C$, $C$ beats $A$) in terms of the $w$'s.

Dissect the task

1. Here is the distinguishing mark of the Olympiad tier: this problem cannot be reached by structure-spotting alone. Two pieces of equipment are needed before it is even fair to pose: the habit of counting one thing in two ways, and fluency with an algebraic identity. In your own school, when would you introduce each, and to whom?
2. The tiling and the sign game were elementary but hard. This one is technical and hard. What is the honest difference in what the child brings to the table?