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


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Question Number 61273 by alphaprime last updated on 31/May/19

$S u p p o s e α, β, γ, δ a r e r e a l n u m b e r s$ $s u c h t h a t α + β + γ + δ = α^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $P r o v e t h a t α (α + β) (α + γ) (α + δ) = 0$
Commented by Rasheed.Sindhi last updated on 31/May/19

$α + β + γ + δ = α^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $It can be proved that any two of$ $four have same value and remaining$ $two are additive inverse of the mentioned$ $two numbers (Only possibility)$ $So$ $Let α = β = k, γ = δ = - k$ $α + β + γ + δ = α^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $\Rightarrow k + k - k - k = k^{7} + k^{7} - k^{7} - k^{7} = 0$ $α (α + β) (α + γ) (α + δ)$ $= k (k + k) (k - k) (k -)$ $= k . 2 k . 0.0 = 0$ $Proved$
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
	$α+β+γ+δ = 0 ..............(i) ∧ α^7 +β^7 +γ^7 +δ^7 =0............(ii) (i)⇒α=−(β+γ+δ) (ii)⇒{−(β+γ+δ)}^7 +β^7 +γ^7 +δ^7 =0 −(β+γ+δ)^7 =−(β^7 +γ^7 +δ^7 ) (β+γ+δ)^7 =β^7 +γ^7 +δ^7 possible solutions: {β,γ,δ}={0,0,0}⇒α=0 {α,β,γ,δ}={0,0,0,0} For β=k≠0 {β,γ,δ}={k,0,0}⇒α=−(k+0+0)=−k {α,β,γ,δ}={k,−k,0,0} (This means all the possible combinations of k,−k,0,0) (m+n−n)^7 =m^7 +n^7 +(−n)^7 ^• When {α,β,γ,δ}={0,0,0,0} α(α+β)(α+γ)(α+δ) =0(0+0)(0+0)(0+0)=0 ^• When {α,β,γ,δ}={k,−k,0,0} α(α+β)(α+γ)(α+γ) =k(k−k)(k+0)(k+0)=0 Note:Different combinations of k,−k,0,0 haave no effect on the result of the above expression. Hence proved$
	$α + β + γ + δ = 0 . . . . . . . . . . . . . . (i)$ $\land$ $α^{7} + β^{7} + γ^{7} + δ^{7} = 0 . . . . . . . . . . . . (ii)$ $(i) \Rightarrow α = - (β + γ + δ)$ $(ii) \Rightarrow {- (β + γ + δ)}^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $- {(β + γ + δ)}^{7} = - (β^{7} + γ^{7} + δ^{7})$ ${(β + γ + δ)}^{7} = β^{7} + γ^{7} + δ^{7}$ $possible solutions :$ ${β, γ, δ} = {0, 0, 0} \Rightarrow α = 0$ ${α, β, γ, δ} = {0, 0, 0, 0}$ $For β = k \neq 0$ ${β, γ, δ} = {k, 0, 0} \Rightarrow α = - (k + 0 + 0) = - k$ ${α, β, γ, δ} = {k, - k, 0, 0}$ $(This means all the possible combinations$ $of k, - k, 0, 0)$ ${(m + n - n)}^{7} = m^{7} + n^{7} + {(- n)}^{7}$ $^{∙} When {α, β, γ, δ} = {0, 0, 0, 0}$ $α (α + β) (α + γ) (α + δ)$ $= 0 (0 + 0) (0 + 0) (0 + 0) = 0$ $^{∙} When {α, β, γ, δ} = {k, - k, 0, 0}$ $α (α + β) (α + γ) (α + γ)$ $= k (k - k) (k + 0) (k + 0) = 0$ $Note : Different combinations of$ $k, - k, 0, 0 haave no effect on$ $the result of the above expression .$ $Hence proved$
	Commented by alphaprime last updated on 02/Jun/19
	Thank you gentlemen :)
	Commented by Rasheed.Sindhi last updated on 02/Jun/19
	$Thanks but the answer is not satisfactory yet! Because It doesn′t covers all the possibilities. For example α=a≠0 , β=b≠0, γ=−a & δ=−b. α+β+γ+δ=0⇒a+b−a−b=0 α^7 +β^7 +γ^7 +δ^7 =0 ⇒a^7 +b^7 +(−a)^7 +(−b)^7 =a^7 +b^7 −a^7 −b^7 =0 α(α+β)(α+γ)(α+δ) =a(a+b)(a−a)(a−b)=0 How the solution set: {α,β,γ,δ}={a,b,−a,−b} can be derived from (β+γ+δ)^7 =β^7 +γ^7 +δ^7 ? I think {α,β,γ,δ}={a,b,−a,−b} where a,b∈R (a,b may be zero also) is complete solution.$
	$Thanks but the answer is not$ $satisfactory yet! Because It {doesn}^{'} t$ $covers all the possibilities .$ $For example α = a \neq 0, β = b \neq 0,$ $γ = - a & δ = - b .$ $α + β + γ + δ = 0 \Rightarrow a + b - a - b = 0$ $α^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $\Rightarrow a^{7} + b^{7} + {(- a)}^{7} + {(- b)}^{7}$ $= a^{7} + b^{7} - a^{7} - b^{7} = 0$ $α (α + β) (α + γ) (α + δ)$ $= a (a + b) (a - a) (a - b) = 0$ $How the solution set :$ ${α, β, γ, δ} = {a, b, - a, - b}$ $can be derived from$ ${(β + γ + δ)}^{7} = β^{7} + γ^{7} + δ^{7} ?$ $I think {α, β, γ, δ} = {a, b, - a, - b}$ $where a, b \in R (a, b may be zero also)$ $is complete solution .$
	Answered by Rasheed.Sindhi last updated on 03/Jun/19
	$α+β+γ+δ =0.............(i) ∧ α^7 +β^7 +γ^7 +δ^7 =0..........(ii) ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢^(−) Let α=a , β=b where a,b∈R (i)⇒γ+δ=−a−b.............(iii) (ii)⇒γ^7 +δ^7 =−a^7 −b^7 .......(iv) (iii) & (iv) have only two real solutions (See the answer & comments of sir MJS to Q#61470) and they are γ=−a ∧ δ=−b OR γ=−b ∧ δ=−a Hence the general solution to α+β+γ+δ = α^7 +β^7 +γ^7 +δ^7 =0 is {α,β,γ,δ}={a,b,−a,−b} for real α,β,γ,δ. To prove α(α+β)(α+γ)(α+δ)=0 α(α+β)(α+γ)(α+δ) =a(a+b)(a−a)(a−b) =a(a+b)(0)(a−b)=0$
	$α + β + γ + δ = 0 . . . . . . . . . . . . . (i)$ $\land$ $α^{7} + β^{7} + γ^{7} + δ^{7} = 0 . . . . . . . . . . (ii)$ $\overset{―}{⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢}$ $Let α = a, β = b where a, b \in R$ $(i) \Rightarrow γ + δ = - a - b . . . . . . . . . . . . . (iii)$ $(ii) \Rightarrow γ^{7} + δ^{7} = - a^{7} - b^{7} . . . . . . . (iv)$ $(iii) & (iv) have only two real solutions$ $(See the answer & comments of sir$ $You can't use 'macro parameter character #' in math mode$ $γ = - a \land δ = - b$ $OR$ $γ = - b \land δ = - a$ $H ence the general solution to$ $α + β + γ + δ = α^{7} + β^{7} + γ^{7} + δ^{7} = 0$ $is {α, β, γ, δ} = {a, b, - a, - b} for real$ $α, β, γ, δ .$ $To prove α (α + β) (α + γ) (α + δ) = 0$ $α (α + β) (α + γ) (α + δ)$ $= a (a + b) (a - a) (a - b)$ $= a (a + b) (0) (a - b) = 0$
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