Each of x, y and z is a rational number such that: x + y * ³√2 + z * ³√4 = 0.

Prove that x=y=z=0 is the only possible solution to the above equation.

I'm using the same strategy as Harry in my solution, with a generalization, let c be any integer which is not a perfect cube (non-zero is covered by not perfect cube).

Given Equation
(1) x + ©ø¡î(c)*y + ©ø¡î(c©÷)*z = 0

Rearrange and square (1)
(2) x©÷ = ©ø¡î(c©÷)*y©÷ + 2c*yz +c*©ø¡îc*z©÷

Equation 2 multiplied by z and rearranged
(3) x©÷*z = (©ø¡î(c©÷)*z)*y©÷ + 2c*yz©÷ +c*©ø¡îc*z©ø

Substitute (1) into (3) to eliminate (©ø¡îc^2*z)
(4) x©÷*z = (-x - ©ø¡îc*y)*y©÷ + 2c*yz©÷ +c*©ø¡îc*z©ø

Rearrange (4) according to terms with cbrt(c)
(5) x©÷*z + y©÷*z - 2*c*y*z©÷ = ©ø¡îc*(c*z©ø - y©ø)

The left side (x©÷*z + y©÷*z - 2*c*y*z©÷) is rational.  The right factor (c*z©ø - y©ø) is rational, but ©ø¡îc is irrational. The only way the equation is true is if the rational factors are zero.

Equations without formatting:
(1) x + cbrt[c]*y + cbrt[c^2]*z = 0
(2) x^2 = cbrt[c^2]*y^2 + 2c*yz +c*cbrt[c]*z^2
(3) x^2*z = (cbrt[c^2]*z)*y^2 + 2c*yz^2 +c*cbrt[c]*z^3
(4) x^2*z = (-x - cbrt[c]*y)*y^2 + 2c*yz^2 +c*cbrt[c]*z^3
(5) x^2*z + y^2*z - 2*c*y*z^2 = cbrt[c]*(c*z^3 - y^3)

Posted by Brian Smith on 2009-10-16 12:04:25