在一个 $2013 \times 2013$ 的正数数表中，每行都成等差数列，每列平方后都成等差数列，求证：左上角的数和右下角的数之积等于左下角的数和右上角的数之积．
【难度】
【出处】
2013年北京大学保送生试题
【标注】
【答案】
略
【解析】
下面证明对 $n \times n$ 的数表，$n\geqslant 3,n \in {\mathbb{N}^*}$，$n$ 是奇数，命题均成立．
当 $n = 2k + 1$ 时，不妨设数表如下．$$\begin{array}{|c|c|c|c|c|}\hline a &\cdots &\dfrac{a+b}{2}&\cdots & b\\ \hline \sqrt {\dfrac{{{a^2} + {c^2}}}{2}}&\cdots &\sqrt {\dfrac{{{{\left( {\dfrac{{a + b}}{2}} \right)}^2} + {{\left( {\dfrac{{c + d}}{2}} \right)}^2}}}{2}} &\cdots & \sqrt {\dfrac{{{b^2} + {d^2}}}{2}} \\ \hline \cdots &\cdots &\cdots &\cdots &\cdots \\ \hline c &\cdots &\dfrac{{c + d}}{2}&\cdots & d\\ \hline \end{array}$$于是$$2\sqrt {\frac{{{{\left( {\dfrac{{a + b}}{2}} \right)}^2} + {{\left( {\dfrac{{c + d}}{2}} \right)}^2}}}{2}} = \sqrt {\dfrac{{{a^2} + {c^2}}}{2}} + \sqrt {\dfrac{{{b^2} + {d^2}}}{2}} ,$$即$${\left( {a + b} \right)^2} + {\left( {c + d} \right)^2} = {a^2} + {c^2} + {b^2} + {d^2} + 2\sqrt {\left( {{a^2} + {c^2}} \right)\left( {{b^2} + {d^2}} \right)} ,$$所以$$ {\left( {ab + cd} \right)^2} = \left( {{a^2} + {c^2}} \right)\left( {{b^2} + {d^2}} \right),$$也即$$ 2abcd = {b^2}{c^2} + {a^2}{d^2},$$因此$$ ad = bc.$$由此命题成立．
当 $n = 2k + 1$ 时，不妨设数表如下．$$\begin{array}{|c|c|c|c|c|}\hline a &\cdots &\dfrac{a+b}{2}&\cdots & b\\ \hline \sqrt {\dfrac{{{a^2} + {c^2}}}{2}}&\cdots &\sqrt {\dfrac{{{{\left( {\dfrac{{a + b}}{2}} \right)}^2} + {{\left( {\dfrac{{c + d}}{2}} \right)}^2}}}{2}} &\cdots & \sqrt {\dfrac{{{b^2} + {d^2}}}{2}} \\ \hline \cdots &\cdots &\cdots &\cdots &\cdots \\ \hline c &\cdots &\dfrac{{c + d}}{2}&\cdots & d\\ \hline \end{array}$$于是$$2\sqrt {\frac{{{{\left( {\dfrac{{a + b}}{2}} \right)}^2} + {{\left( {\dfrac{{c + d}}{2}} \right)}^2}}}{2}} = \sqrt {\dfrac{{{a^2} + {c^2}}}{2}} + \sqrt {\dfrac{{{b^2} + {d^2}}}{2}} ,$$即$${\left( {a + b} \right)^2} + {\left( {c + d} \right)^2} = {a^2} + {c^2} + {b^2} + {d^2} + 2\sqrt {\left( {{a^2} + {c^2}} \right)\left( {{b^2} + {d^2}} \right)} ,$$所以$$ {\left( {ab + cd} \right)^2} = \left( {{a^2} + {c^2}} \right)\left( {{b^2} + {d^2}} \right),$$也即$$ 2abcd = {b^2}{c^2} + {a^2}{d^2},$$因此$$ ad = bc.$$由此命题成立．
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