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Question Number 70135 by petrochengula last updated on 01/Oct/19
sophie−Germain identity  a^4 +4b^4 =((a+b)^2 +b^2 )((a−b)^2 +b^2 )
$${sophie}−{Germain}\:{identity} \\ $$$${a}^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{4}} =\left(\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\left({a}−{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$
Commented by mathmax by abdo last updated on 02/Oct/19
{(a+b)^2 +b^2 ){(a−b)^2  +b^2 }=  =(a^2 +2ab +2b^2 )(a^2 −2ab +2b^2 )  =(a^2 +2b^2 +2ab)(a^2 +2b^2 −2ab)  =(a^2 +2b^2 )^2 −4a^2 b^2 =a^4 +4a^2 b^2  +4b^4 −4a^2 b^2   =a^4  +4b^4
$$\left\{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left\{\left({a}−{b}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right\}= \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{2}{ab}\:+\mathrm{2}{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} −\mathrm{2}{ab}\:+\mathrm{2}{b}^{\mathrm{2}} \right) \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}\right)\left({a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ab}\right) \\ $$$$=\left({a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} ={a}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:+\mathrm{4}{b}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{4}} \:+\mathrm{4}{b}^{\mathrm{4}} \\ $$

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