X1+y1= x2+y2 = x3+y3 = ... = X6+y6 = 5 (each person plays one game with five other people, ***5 games).

By transferring the above conclusions, we can get:

(x 1^2-y 1^2)+……+(x6^2+y6^2)=0.

(x 1+y 1)(x 1-y 1)+……+(X6+Y6)(X6-Y6)= 0

5x 1-5y 1+……+5x 6+5y 6 = 0

5(x 1+X2+X3+X4+X5+X6)-5(y 1+Y2+Y3+Y4+Y5+Y6)= 0

And because there are winners and losers, one person wins and one person loses, so the sum of x = the sum of y,

So left =0= right above, so it holds.

Give me some hard work.