How to Solve a Radical Equation with Golden Flavor

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[wes]

Because of symmetry in x^3 = 2y – 1 and y^3 = 2x – 1, valid solutions satisfy y = x.
So set x^3 – 2x +1 = 0. x = 1 is a solution and synthetic division by (x – 1) gives x^2 + x – 1 = 0.
The quadratic solution to this equation is x = (–1 ± √5)/2
Let Φ = (√5 + 1)/2 be the Golden Ratio and its inverse φ = 1/Φ = Φ – 1 = (√5 – 1)/2
Thus, x = y = φ = (√5 – 1)/2 is a possible solution
Check:
φ = Φ – 1 = 1/Φ
φ^2 = φ(Φ – 1) = 1 – φ
φ^3 = φ(1 – φ) = φ – φ^2 = 2φ – 1
Thus x^3 + 1 = φ^3 + 1 = 2φ
(x^3 + 1)^3 = 8φ^3 = 8( 2φ – 1) = 8(2x – 1) = [2(2x – 1)^(1/3)]^3
Thus, x = y = φ = (√5 – 1)/2

Is x = y = –(√5 + 1)/2 = –Φ also a solution?
Φ^3 = 2Φ + 1 so x^3 = (–Φ)^3 = –Φ^3 = –(2Φ + 1) and x^3 + 1 = –2Φ
Thus, (x^3 + 1)^3 = –8Φ^3 = 16Φ
On the right hand side we have 8(2x – 1) = 8(–2Φ – 1), which is unequal to 16Φ
Thus, x = y = –Φ is not a solution.

[yoav]

Nice. Only (sqrt5-1)/2 is a solution, (-sqrt5-1)/2 dose not work.

[grink_man]

I want to try to solve this equation using the nonic equation.
x⁹+3x⁶+3x³-16x+9=0
1+3+3-16+9=1+6+9-16=16-16=0, so x=1. Divide the nonic by (x-1):
Let's try to find the root among the divisors of the free term: ±1, ±3, ±9.
x=1: 1+1+1+4+4+4+7+7-9=3+12+14-9=29-9=20, so x cannot be more than 0.
x=-1: 1-1+1-4+4-4+7-7-9=-12, so x cannot be less than 0. But also x cannot be equal to 0. Hmm... This may mean that the equation has no other roots at all... or only other roots ∈ Z.

Let's try to factorize it by the other way.
x⁶(x²+x+1)+4x³(x²+x+1)+7(x²+x+1)-16=(x²+x+1)(x⁶+4x³+7)-16=0, (x²+x+1)(x⁶+4x³+7)=16. x²+x+1 and x⁶+4x³+7 are both more than 0, so ab=1*16=2*8=4*4=8*2=16*1, where a=x²+x+1, b=x⁶+4x³+7

1) x²+x+1=1, x⁶+4x³+7=16.
x²+x=0, so x=0; -1. 7≠16 - false; 1-4+7=8-4=4≠16 - false. So a≠1, b≠16.
2) x²+x+1=2, x⁶+4x³+7=8
x²+x-1=0. D=1-4*(-1)=1+4=5>0. x1, 2=(-1±sqrt(5))/2.
x⁶+4x³-1 is triquadratic. x³=t, so x⁶=t². t²+4t-1=0, D(t)=16-4*(-1)=16+4=20>0. t1, 2=(-4±sqrt(20))/2=(-2±sqrt(5)). (-(1/2)*(1±sqrt(5)))³=-(1/8)*(1±3sqrt(5)+3*5±5sqrt(5))=-(1/8)(16±8sqrt(5))=(-2±sqrt(5)), so t1, 2=(x1, 2)³. So, x=(-1±sqrt(5))/2 are roots of the equation.
3) x²+x+1=4, x⁶+4x³+7=4.
x²+x-3=0. D=1-4*(-3)=1+12=13>0. x3, 4=(-1±sqrt(13))/2. x⁶+4x³+3=0; t²+4t+3=0. 1-4+3=0, so t=-1 is the root. (t+1)(t+3)=0, t=-1; -3. x=cbrt(-1)=-1; x=cbrt(-3). So these numbers aren't the roots of the equation.
4) x²+x+1=8, x⁶+4x³+7=2.
x²+x-7=0. D=1+4*7=1+28=29>0. x3, 4=(-1±sqrt(29))/2. x⁶+4x³+5=0; t²+4t+5=0. D(t)=16-4*5=16-20=-4<0, so x⁶+4x³+5>0. No roots
5) x²+x+1=16, x⁶+4x³+7=1; x⁶+4x³+6>x⁶+4x³+5>0. No roots.

By solving this absurdly hard equation I found 3 rational roots: 1; (sqrt(5)-1)/2; -(sqrt(5)+1)/2