Suppose-are-real-numbers-such-that-7-7-7-7-0-Prove-that-0-

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 \dots \dots \dots \dots . . (i)

\land

α^{7} + β^{7} + γ^{7} + δ^{7} = 0 \dots \dots \dots \dots (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 \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 \dots \dots \dots \dots . (i)

\land

α^{7} + β^{7} + γ^{7} + δ^{7} = 0 \dots \dots \dots . (ii)

\overset{―}{⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢}

Let α = a, β = b where a, b \in R

(i) \Rightarrow γ + δ = - a - b \dots \dots \dots \dots . (iii)

(ii) \Rightarrow γ^{7} + δ^{7} = - a^{7} - b^{7} \dots \dots . (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