99松山高中
例題 已知
146 = {5^2} + {11^2},
218 = {7^2} + {13^2},試將
146 \times 218 = 31828
表示成兩個正整數的平方和?
解:
146 \times 218 = {\left( t \right)^2} + {\left( k \right)^2} t是正整數,
k是正整數。
(1) 令
a = 5,b = 11,c = 7,d = 13
\begin{align}
& 146\times 218=\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right) \\
& ={{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}} \\
& =\left\{ {{\left( ac \right)}^{2}}+2\left( ac \right)\left( bd \right)+{{\left( bd \right)}^{2}} \right\}+\left\{ {{\left( ad \right)}^{2}}+2\left( ad \right)\left( bc \right)+{{\left( bc \right)}^{2}} \right\} \\
& ={{\left\{ ac+bd \right\}}^{2}}+{{\left\{ ad-bc \right\}}^{2}} \\
& ={{\left( 5\times 7+11\times 13 \right)}^{2}}+{{\left( 5\times 13-11\times 7 \right)}^{2}} \\
& ={{178}^{2}}+{{\left( -12 \right)}^{2}} \\
\end{align}
答案
178 與
12
想法: 順著題目意思,把
5,11,7,13改成
a,b,c,d,比較好運算,接著配成完全平方。