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Question Number 109544 by 1549442205PVT last updated on 25/Aug/20

Exclude m and n from the equalities: a=m+n,b^3 =m^3 +n^3 ,c^5 =m^5 +n^5

$$\mathrm{Exclude}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{from}\:\mathrm{the}\:\mathrm{equalities}: \\ $$$$\mathrm{a}=\mathrm{m}+\mathrm{n},\mathrm{b}^{\mathrm{3}} =\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} ,\mathrm{c}^{\mathrm{5}} =\mathrm{m}^{\mathrm{5}} +\mathrm{n}^{\mathrm{5}} \\ $$

Commented by Her_Majesty last updated on 24/Aug/20

only 2 equalities are necessary to get m and n ⇒ it′s an easy task let m=x−y∧n=x+y a=2x ⇒ x=(a/2) b^3 =3ay^2 +(a^3 /4) ⇒ y^2 =((4b^3 −a^3 )/(12a)) c^5 =((5b^6 +5a^3 b^3 −a^6 )/(9a))

$${only}\:\mathrm{2}\:{equalities}\:{are}\:{necessary}\:{to}\:{get}\:{m}\:{and} \\ $$$${n}\:\Rightarrow\:{it}'{s}\:{an}\:{easy}\:{task} \\ $$$${let}\:{m}={x}−{y}\wedge{n}={x}+{y} \\ $$$${a}=\mathrm{2}{x}\:\Rightarrow\:{x}=\frac{{a}}{\mathrm{2}} \\ $$$${b}^{\mathrm{3}} =\mathrm{3}{ay}^{\mathrm{2}} +\frac{{a}^{\mathrm{3}} }{\mathrm{4}}\:\Rightarrow\:{y}^{\mathrm{2}} =\frac{\mathrm{4}{b}^{\mathrm{3}} −{a}^{\mathrm{3}} }{\mathrm{12}{a}} \\ $$$${c}^{\mathrm{5}} =\frac{\mathrm{5}{b}^{\mathrm{6}} +\mathrm{5}{a}^{\mathrm{3}} {b}^{\mathrm{3}} −{a}^{\mathrm{6}} }{\mathrm{9}{a}} \\ $$

Answered by ajfour last updated on 24/Aug/20

m^3 +n^3 =(m+n){(m+n)^2 −3mn} b^3 =a(a^2 −3mn) m^5 +n^5 =(m^3 +n^3 )(m^2 +n^2 ) −m^2 n^2 (m+n) ⇒ c^5 =b^3 {a^2 −(2/3)(a^2 −(b^3 /a))}−(a/9)(a^2 −(b^3 /a))^2

$${m}^{\mathrm{3}} +{n}^{\mathrm{3}} =\left({m}+{n}\right)\left\{\left({m}+{n}\right)^{\mathrm{2}} −\mathrm{3}{mn}\right\} \\ $$$${b}^{\mathrm{3}} ={a}\left({a}^{\mathrm{2}} −\mathrm{3}{mn}\right) \\ $$$${m}^{\mathrm{5}} +{n}^{\mathrm{5}} =\left({m}^{\mathrm{3}} +{n}^{\mathrm{3}} \right)\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{m}^{\mathrm{2}} {n}^{\mathrm{2}} \left({m}+{n}\right)\:\:\: \\ $$$$\Rightarrow \\ $$$$\:{c}^{\mathrm{5}} ={b}^{\mathrm{3}} \left\{{a}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\left({a}^{\mathrm{2}} −\frac{{b}^{\mathrm{3}} }{{a}}\right)\right\}−\frac{{a}}{\mathrm{9}}\left({a}^{\mathrm{2}} −\frac{{b}^{\mathrm{3}} }{{a}}\right)^{\mathrm{2}} \\ $$

Commented by 1549442205PVT last updated on 25/Aug/20

Thank both Sirs

$$\mathrm{Thank}\:\mathrm{both}\:\mathrm{Sirs} \\ $$

Answered by 1549442205PVT last updated on 25/Aug/20

From the hypothesis we have a^3 =(m+n)^3 =m^3 +n^3 +3mn(m+n)= b^3 +3mna⇒mn=((a^3 −b^3 )/(3a))(1) a^5 =(m+n)^5 =m^5 +n^5 +5mn(m^3 +n^3 ) +10(mn)^2 (m+n)=c^5 +5mnb^3 + +10mna(2).Replace(1)into (2) we get a^5 −c^5 =5b^3 .((a^3 −b^3 )/(3a))+10a.(((a^3 −b^3 )/(3a)))^2 =((a^3 −b^3 )/(3a))(5b^3 +10a.((a^3 −b^3 )/(3a))) =((a^3 −b^3 )/(3a)).((10a^3 +5b^3 )/3).Finally,we obtain 9a(a^5 −c^5 )=5(a^3 −b^3 )(2a^3 +b^3 )

$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{a}^{\mathrm{3}} =\left(\mathrm{m}+\mathrm{n}\right)^{\mathrm{3}} =\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} +\mathrm{3mn}\left(\mathrm{m}+\mathrm{n}\right)= \\ $$$$\mathrm{b}^{\mathrm{3}} +\mathrm{3mna}\Rightarrow\mathrm{mn}=\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}\left(\mathrm{1}\right) \\ $$$$\mathrm{a}^{\mathrm{5}} =\left(\mathrm{m}+\mathrm{n}\right)^{\mathrm{5}} =\mathrm{m}^{\mathrm{5}} +\mathrm{n}^{\mathrm{5}} +\mathrm{5mn}\left(\mathrm{m}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} \right) \\ $$$$+\mathrm{10}\left(\mathrm{mn}\right)^{\mathrm{2}} \left(\mathrm{m}+\mathrm{n}\right)=\mathrm{c}^{\mathrm{5}} +\mathrm{5mnb}^{\mathrm{3}} + \\ $$$$+\mathrm{10mna}\left(\mathrm{2}\right).\mathrm{Replace}\left(\mathrm{1}\right)\mathrm{into}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{5}} −\boldsymbol{\mathrm{c}}^{\mathrm{5}} =\mathrm{5}\boldsymbol{\mathrm{b}}^{\mathrm{3}} .\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}+\mathrm{10a}.\left(\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}\left(\mathrm{5b}^{\mathrm{3}} +\mathrm{10a}.\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}\right) \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} }{\mathrm{3a}}.\frac{\mathrm{10a}^{\mathrm{3}} +\mathrm{5b}^{\mathrm{3}} }{\mathrm{3}}.\mathrm{Finally},\mathrm{we}\:\mathrm{obtain} \\ $$$$\mathrm{9}\boldsymbol{\mathrm{a}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{5}} −\boldsymbol{\mathrm{c}}^{\mathrm{5}} \right)=\mathrm{5}\left(\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{b}}^{\mathrm{3}} \right)\left(\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} \right) \\ $$

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