| Version current |
Version 22 |
| \begin{thmplain} |
\begin{thmplain} |
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\,\, The set of the sums of two squares of integers is closed under multiplication; in fact we have the identical equation
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\,\, The set of the sums of two squares of integers is \PMlinkescapetext{closed} under multiplication; in fact we have the identical equation
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$$(a^2+b^2)(c^2+d^2) \;=\; (ac-bd)^2+(ad+bc)^2.$$
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$$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2.$$
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| \end{thmplain} |
\end{thmplain} |
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| This was presented by Leonardo Fibonacci in 1225 (in ``Liber quadratorum''), but the original inventor was Brahmagupta. |
This was presented by Leonardo Fibonacci in 1225 (in ``Liber quadratorum''), but the original inventor was Brahmagupta. |
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| The proof of the equation may utilize imaginary numbers as follows: |
The proof of the equation may utilize imaginary numbers as follows: |
| \begin{align*} |
\begin{align*} |
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(a^2+b^2)(c^2+d^2) & \;=\; (a+ib)(a-ib)(c+id)(c-id)\\
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(a^2+b^2)(c^2+d^2) & = (a+ib)(a-ib)(c+id)(c-id)\\
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& \;=\; (a+ib)(c+id)(a-ib)(c-id)\\
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& = (a+ib)(c+id)(a-ib)(c-id)\\
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& \;=\; [(ac-bd)+i(ad+bc)][(ac-bd)-i(ad+bc)]\\
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& = [(ac-bd)+i(ad+bc)][(ac-bd)-i(ad+bc)]\\
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& \;=\; (ac-bd)^2+(ad+bc)^2
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& = (ac-bd)^2+(ad+bc)^2
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| \end{align*} |
\end{align*} |
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| \textbf{Note.}\, The equation is the special case\, $n = 2$\, of Lagrange's identity. |
\textbf{Note.}\, The equation is the special case\, $n = 2$\, of Lagrange's identity. |
