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I was given this problem and I think I solved it but I wanted to double-check my solution using the collective wisdom here. Here is the problem:

Twelve teams play in a round-robin tournament. Each match results in a win or loss (no ties). If the team $n$ wins $a_n$ matches and loses $b_n$ matches, prove that $$\sum_{n=1}^{12} a_n^2 = \sum_{n=1}^{12}{b_n^2}$$

My work: the number of all matches is 66. So the total number of wins and losses will be 66. $$\sum_{n=1}^{12} a_n = \sum_{n=1}^{12}{b_n}=66$$ It's also easy to see that $b_n=11-a_n$, thus $$\sum_{n=1}^{12}{b_n^2}=\sum_{n=1}^{12}{(121 -22a_n +a_n^2)}=1452 - 22\sum_{n=1}^{12}{a_n} + \sum_{n=1}^{12}{a_n^2}= \sum_{n=1}^{12}{a_n^2}$$

Is there an easier solution? Thanks in advance!

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  • $\begingroup$ Looks right to me. $\endgroup$
    – Mosquite
    Aug 25 '17 at 20:50
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Perhaps slightly simpler (or at least a little better looking) is this . . .

\begin{align*} &\sum a_n^2 - \sum b_n^2\\[4pt] =\;&\sum (a_n^2 - b_n^2)\\[4pt] =\;&\sum (a_n+b_n)(a_n-b_n)\\[4pt] =\;&\sum 11(a_n-b_n)\\[4pt] =\;\,&11\sum (a_n-b_n)\\[4pt] =\;\,&11\left(\sum a_n- \sum b_n\right)\\[4pt] =\;\,&11(0)\\[4pt] =\;\,&0\\[4pt] \end{align*}

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