# 婆罗摩笈多-斐波那契恒等式

{\displaystyle {\begin{aligned}\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)&{}=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2}\ \qquad \qquad (1)\\&{}=\left(ac+bd\right)^{2}+\left(ad-bc\right)^{2}.\qquad \qquad (2)\end{aligned}}}

${\displaystyle (1^{2}+4^{2})(2^{2}+7^{2})=30^{2}+1^{2}=26^{2}+15^{2}.\,}$

(1)和(2)都可以用展开多项式的方法来证实。(2)可以通过把(1)中的${\displaystyle b}$换成${\displaystyle -b}$来得出。

## 證明

{\displaystyle {\begin{aligned}\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(a^{2}c^{2}+b^{2}d^{2}{\color {red}-2abcd}\right)+\left(a^{2}d^{2}+b^{2}c^{2}{\color {red}+2abcd}\right)\\&=\left(ac-bd\right)^{2}+\left(ad+bc\right)^{2}\end{aligned}}}

${\displaystyle \left(ac+bd\right)^{2}+\left(ad-bc\right)^{2}}$

## 与复数的关系

${\displaystyle |a+bi||c+di|=|(a+bi)(c+di)|\,}$

${\displaystyle |a+bi||c+di|=|(ac-bd)+i(ad+bc)|,\,}$

${\displaystyle |a+bi|^{2}|c+di|^{2}=|(ac-bd)+i(ad+bc)|^{2},\,}$

${\displaystyle (a^{2}+b^{2})(c^{2}+d^{2})=(ac-bd)^{2}+(ad+bc)^{2}.\,}$

## 用范数来解释

${\displaystyle a}$ ${\displaystyle b}$ ${\displaystyle c}$ ${\displaystyle d}$ 有理数的情况中，这个等式可以解释为${\displaystyle Q(i)}$ 范数是积性的。也就是说：

${\displaystyle N(a+bi)=a^{2}+b^{2}\,}$ ${\displaystyle N(c+di)=c^{2}+d^{2},\,}$

${\displaystyle N[(a+bi)(c+di)]=N[(ac-bd)+i(ad+bc)]=(ac-bd)^{2}+(ad+bc)^{2}.\,}$

${\displaystyle N[(a+bi)(c+di)]=N(a+bi)\cdot N(c+di).\,}$