Find answers of this system of equations in reals {x+3y=4y^3, y+3z=4z^3, z+3x=4x^3

David Ali

David Ali

Answered question

2022-09-20

Find answers of this system of equations in reals
{ x + 3 y = 4 y 3 y + 3 z = 4 z 3 z + 3 x = 4 x 3

Answer & Explanation

Nathalie Rivers

Nathalie Rivers

Beginner2022-09-21Added 7 answers

Suppose that we had x > 1. Then, since 4 x 3 3 x > x, we have z > x, and similarly y > z, x > y, contradiction. By symmetry, we conclude that x , y , z [ 1 , 1 ].
So there exist α , β , γ [ 0 , π ] with x = cos α, y = cos β, z = cos γ. By the formula for cos 3 α, we can rewrite the system of equations as:
{ α ± 3 β   ( mod   2 π ) β = ± 3 γ   ( mod   2 π ) γ = ± 3 α   ( mod   2 π )
So we have ± 27 α α, so either 26 α 0 or 28 α 0. We find that α = π k / 13 or α = π k / 14 for some nonnegative integer k.
This gives 27 solutions, x = cos π k 13 for 0 k 13, and x = cos π k 14 for 1 k 13.
For example, one solution is ( ( cos π 14 , cos 9 π 14 , cos 3 π 14 )).

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