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Question Number 61273 by alphaprime last updated on 31/May/19
Suppose α ,β,γ,δ are real numbers such that α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 Prove that α(α+β)(α+γ)(α+δ)=0
$${Suppose}\:\alpha\:,\beta,\gamma,\delta\:{are}\:{real}\:{numbers} \\ $$$${such}\:{that}\:\alpha+\beta+\gamma+\delta\:=\:\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$${Prove}\:{that}\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right)=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 31/May/19
α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 It can be proved that any two of four have same value and remaining two are additive inverse of the mentioned two numbers(Only possibility) So Let α=β=k ,γ=δ=−k α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 ⇒ k+k−k−k=k^7 +k^7 −k^7 −k^7 =0 α(α+β)(α+γ)(α+δ) =k(k+k)(k−k)(k−) =k.2k.0.0=0 Proved
$$\:\:\:\alpha+\beta+\gamma+\delta\:=\:\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$$\mathrm{It}\:\mathrm{can}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{that}\:\mathrm{any}\:\mathrm{two}\:\mathrm{of} \\ $$$$\mathrm{four}\:\mathrm{have}\:\mathrm{same}\:\mathrm{value}\:\mathrm{and}\:\mathrm{remaining} \\ $$$$\mathrm{two}\:\mathrm{are}\:\mathrm{additive}\:\mathrm{inverse}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mentioned} \\ $$$$\mathrm{two}\:\mathrm{numbers}\left(\mathrm{Only}\:\mathrm{possibility}\right) \\ $$$$\mathrm{So} \\ $$$$\mathrm{Let}\:\:\alpha=\beta=\mathrm{k}\:,\gamma=\delta=−\mathrm{k} \\ $$$$\:\:\:\alpha+\beta+\gamma+\delta\:=\:\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\mathrm{k}+\mathrm{k}−\mathrm{k}−\mathrm{k}=\mathrm{k}^{\mathrm{7}} +\mathrm{k}^{\mathrm{7}} −\mathrm{k}^{\mathrm{7}} −\mathrm{k}^{\mathrm{7}} =\mathrm{0} \\ $$$$\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right) \\ $$$$\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{k}+\mathrm{k}\right)\left(\mathrm{k}−\mathrm{k}\right)\left(\mathrm{k}−\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{k}.\mathrm{2k}.\mathrm{0}.\mathrm{0}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{Proved} \\ $$$$ \\ $$
Commented by alphaprime last updated on 01/Jun/19
Could it have been better please?!
Answered by Rasheed.Sindhi last updated on 02/Jun/19
α+β+γ+δ = 0 ..............(i) ∧ α^7 +β^7 +γ^7 +δ^7 =0............(ii) (i)⇒α=−(β+γ+δ) (ii)⇒{−(β+γ+δ)}^7 +β^7 +γ^7 +δ^7 =0 −(β+γ+δ)^7 =−(β^7 +γ^7 +δ^7 ) (β+γ+δ)^7 =β^7 +γ^7 +δ^7 possible solutions: {β,γ,δ}={0,0,0}⇒α=0 {α,β,γ,δ}={0,0,0,0} For β=k≠0 {β,γ,δ}={k,0,0}⇒α=−(k+0+0)=−k {α,β,γ,δ}={k,−k,0,0} (This means all the possible combinations of k,−k,0,0) (m+n−n)^7 =m^7 +n^7 +(−n)^7 ^• When {α,β,γ,δ}={0,0,0,0} α(α+β)(α+γ)(α+δ) =0(0+0)(0+0)(0+0)=0 ^• When {α,β,γ,δ}={k,−k,0,0} α(α+β)(α+γ)(α+γ) =k(k−k)(k+0)(k+0)=0 Note:Different combinations of k,−k,0,0 haave no effect on the result of the above expression. Hence proved
$$\alpha+\beta+\gamma+\delta\:=\:\mathrm{0}\:…………..\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\wedge \\ $$$$\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0}…………\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\Rightarrow\alpha=−\left(\beta+\gamma+\delta\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\left\{−\left(\beta+\gamma+\delta\right)\right\}^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$$−\left(\beta+\gamma+\delta\right)^{\mathrm{7}} =−\left(\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} \right) \\ $$$$\:\:\:\left(\beta+\gamma+\delta\right)^{\mathrm{7}} =\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} \\ $$$$\mathrm{possible}\:\:\mathrm{solutions}: \\ $$$$\:\:\:\left\{\beta,\gamma,\delta\right\}=\left\{\mathrm{0},\mathrm{0},\mathrm{0}\right\}\Rightarrow\alpha=\mathrm{0} \\ $$$$\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right\} \\ $$$$\:\:\:\mathrm{For}\:\beta=\mathrm{k}\neq\mathrm{0} \\ $$$$\:\:\:\:\left\{\beta,\gamma,\delta\right\}=\left\{\mathrm{k},\mathrm{0},\mathrm{0}\right\}\Rightarrow\alpha=−\left(\mathrm{k}+\mathrm{0}+\mathrm{0}\right)=−\mathrm{k} \\ $$$$\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{k},−\mathrm{k},\mathrm{0},\mathrm{0}\right\} \\ $$$$\left(\mathrm{This}\:\mathrm{means}\:\mathrm{all}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{combinations}\right. \\ $$$$\left.\mathrm{of}\:\:\:\mathrm{k},−\mathrm{k},\mathrm{0},\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\left(\mathrm{m}+\mathrm{n}−\mathrm{n}\right)^{\mathrm{7}} =\mathrm{m}^{\mathrm{7}} +\mathrm{n}^{\mathrm{7}} +\left(−\mathrm{n}\right)^{\mathrm{7}} \\ $$$$\:^{\bullet} \mathrm{When}\:\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{0},\mathrm{0},\mathrm{0},\mathrm{0}\right\} \\ $$$$\:\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right) \\ $$$$\:\:\:\:=\mathrm{0}\left(\mathrm{0}+\mathrm{0}\right)\left(\mathrm{0}+\mathrm{0}\right)\left(\mathrm{0}+\mathrm{0}\right)=\mathrm{0} \\ $$$$\:^{\bullet} \mathrm{When}\:\:\:\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{k},−\mathrm{k},\mathrm{0},\mathrm{0}\right\} \\ $$$$\:\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\gamma\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{k}\left(\mathrm{k}−\mathrm{k}\right)\left(\mathrm{k}+\mathrm{0}\right)\left(\mathrm{k}+\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{Note}:\mathrm{Different}\:\mathrm{combinations}\:\mathrm{of} \\ $$$$\:\mathrm{k},−\mathrm{k},\mathrm{0},\mathrm{0}\:\mathrm{haave}\:\mathrm{no}\:\mathrm{effect}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{result}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above}\:\mathrm{expression}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{Hence}\:\mathrm{proved} \\ $$$$\:\: \\ $$
Commented by alphaprime last updated on 02/Jun/19
Thank you gentlemen :)
Commented by Rasheed.Sindhi last updated on 02/Jun/19
Thanks but the answer is not satisfactory yet! Because It doesn′t covers all the possibilities. For example α=a≠0 , β=b≠0, γ=−a & δ=−b. α+β+γ+δ=0⇒a+b−a−b=0 α^7 +β^7 +γ^7 +δ^7 =0 ⇒a^7 +b^7 +(−a)^7 +(−b)^7 =a^7 +b^7 −a^7 −b^7 =0 α(α+β)(α+γ)(α+δ) =a(a+b)(a−a)(a−b)=0 How the solution set: {α,β,γ,δ}={a,b,−a,−b} can be derived from (β+γ+δ)^7 =β^7 +γ^7 +δ^7 ? I think {α,β,γ,δ}={a,b,−a,−b} where a,b∈R (a,b may be zero also) is complete solution.
$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{satisfactory}\:\mathrm{yet}!\:\mathrm{Because}\:\mathrm{It}\:\mathrm{doesn}'\mathrm{t} \\ $$$$\mathrm{covers}\:\mathrm{all}\:\mathrm{the}\:\mathrm{possibilities}. \\ $$$$\mathrm{For}\:\mathrm{example}\:\alpha=\mathrm{a}\neq\mathrm{0}\:,\:\beta=\mathrm{b}\neq\mathrm{0}, \\ $$$$\gamma=−\mathrm{a}\:\&\:\delta=−\mathrm{b}. \\ $$$$\:\alpha+\beta+\gamma+\delta=\mathrm{0}\Rightarrow\mathrm{a}+\mathrm{b}−\mathrm{a}−\mathrm{b}=\mathrm{0} \\ $$$$\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\left(−\mathrm{a}\right)^{\mathrm{7}} +\left(−\mathrm{b}\right)^{\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} −\mathrm{a}^{\mathrm{7}} −\mathrm{b}^{\mathrm{7}} =\mathrm{0} \\ $$$$ \\ $$$$\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right) \\ $$$$\:\:\:\:=\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}−\mathrm{a}\right)\left(\mathrm{a}−\mathrm{b}\right)=\mathrm{0} \\ $$$$\mathrm{How}\:\mathrm{the}\:\:\mathrm{solution}\:\mathrm{set}: \\ $$$$\:\:\:\:\:\:\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{a},\mathrm{b},−\mathrm{a},−\mathrm{b}\right\} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{derived}\:\mathrm{from} \\ $$$$\:\:\:\:\:\:\left(\beta+\gamma+\delta\right)^{\mathrm{7}} =\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} \:\:? \\ $$$$\mathrm{I}\:\mathrm{think}\:\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{a},\mathrm{b},−\mathrm{a},−\mathrm{b}\right\} \\ $$$$\mathrm{where}\:\mathrm{a},\mathrm{b}\in\mathbb{R}\:\left(\mathrm{a},\mathrm{b}\:\mathrm{may}\:\mathrm{be}\:\mathrm{zero}\:\mathrm{also}\right) \\ $$$$\mathrm{is}\:\mathrm{complete}\:\mathrm{solution}. \\ $$
Answered by Rasheed.Sindhi last updated on 03/Jun/19
α+β+γ+δ =0.............(i) ∧ α^7 +β^7 +γ^7 +δ^7 =0..........(ii) ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢^(−) Let α=a , β=b where a,b∈R (i)⇒γ+δ=−a−b.............(iii) (ii)⇒γ^7 +δ^7 =−a^7 −b^7 .......(iv) (iii) & (iv) have only two real solutions (See the answer & comments of sir MJS to Q#61470) and they are γ=−a ∧ δ=−b OR γ=−b ∧ δ=−a Hence the general solution to α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 is {α,β,γ,δ}={a,b,−a,−b} for real α,β,γ,δ. To prove α(α+β)(α+γ)(α+δ)=0 α(α+β)(α+γ)(α+δ) =a(a+b)(a−a)(a−b) =a(a+b)(0)(a−b)=0
$$\alpha+\beta+\gamma+\delta\:=\mathrm{0}………….\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\wedge \\ $$$$\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0}……….\left(\mathrm{ii}\right) \\ $$$$\overline {\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown} \\ $$$$\mathrm{Let}\:\alpha=\mathrm{a}\:,\:\beta=\mathrm{b}\:\mathrm{where}\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\left(\mathrm{i}\right)\Rightarrow\gamma+\delta=−\mathrm{a}−\mathrm{b}………….\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =−\mathrm{a}^{\mathrm{7}} −\mathrm{b}^{\mathrm{7}} …….\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iii}\right)\:\&\:\left(\mathrm{iv}\right)\:\mathrm{have}\:\mathrm{only}\:\mathrm{two}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\left(\mathrm{See}\:\mathrm{the}\:\mathrm{answer}\:\&\:\mathrm{comments}\:\mathrm{of}\:\mathrm{sir}\right. \\ $$$$\left.\:\mathrm{MJS}\:\mathrm{to}\:\mathrm{Q}#\mathrm{61470}\right)\:\mathrm{and}\:\mathrm{they}\:\mathrm{are}\: \\ $$$$\:\:\:\:\:\:\:\:\:\gamma=−\mathrm{a}\:\wedge\:\delta=−\mathrm{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\:\:\:\:\:\:\gamma=−\mathrm{b}\:\wedge\:\delta=−\mathrm{a} \\ $$$$\mathcal{H}\mathrm{ence}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to} \\ $$$$\alpha+\beta+\gamma+\delta\:=\:\alpha^{\mathrm{7}} +\beta^{\mathrm{7}} +\gamma^{\mathrm{7}} +\delta^{\mathrm{7}} =\mathrm{0} \\ $$$$\mathrm{is}\:\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{\mathrm{a},\mathrm{b},−\mathrm{a},−\mathrm{b}\right\}\:\mathrm{for}\:\mathrm{real} \\ $$$$\alpha,\beta,\gamma,\delta. \\ $$$$\: \\ $$$$\mathrm{To}\:\mathrm{prove}\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right)=\mathrm{0} \\ $$$$\:\:\:\alpha\left(\alpha+\beta\right)\left(\alpha+\gamma\right)\left(\alpha+\delta\right) \\ $$$$\:\:\:\:\:=\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}−\mathrm{a}\right)\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\:\:\:\:\:=\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{0}\right)\left(\mathrm{a}−\mathrm{b}\right)=\mathrm{0} \\ $$

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