Suppose, there are $k$ teams playing a round robin tournament that is, each team plays against all other teams and no game ends in a draw. Suppose the $ith$ team loses $l_i$ games and wins $w_i$ games. Show that $$\sum _{i=1}^k l_i^2=\sum_{i=1}^kw_i^2.$$
My solution goes like this:
There are $k$ teams and thus, each team plays $k-1$ games. If the $ith$ wins $w_i$ times, then, $l_i+w_i=k-1$, which implies $ \sum_{i=1}^k(l_i+w_i)=\frac{k(k-1)}{2}.$ Now, $w_i+l_i=k-1$ and thus, $w_i^2-l_i^2=(w_i-l_i)(w_i+l_i)$, due to which $$\sum_{i=1}^k (w_i^2-l_i^2)=\sum_{i=1}^k(l_i+w_i)(w_i-l_i)=(k-1)\sum_{i=1}^k(w_i-l_i).$$ As, no game ends in a tie, thus, $\sum_{i=1}^kw_i=\sum_{i=1}^kl_i. $ More, specifically, the reason can also be stated in a beautiful manner. I got to know, about this reasoning from @leslie townes. The reason goes like this : For each game, we give, one apple to the winning team, and one orange, to the loosing team. No game, ends in a tie. So, for each game, we need to giveaway one apple and one orange. When, the tournament ends, the number of apples given away is equal to the number of oranges which implies the total number of winning matches is equal to the total number of loosing matches. Thus, $\sum_{i=1}^kw_i=\sum_{i=1}^kl_i,$ whence, we conclude, $$\sum_{i=1}^k (w_i^2-l_i^2)=0\implies \sum _{i=1}^k l_i^2=\sum_{i=1}^kw_i^2.$$
Is the solution correct? If not, where is it going wrong?