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x + y * 3√2 + z * 3√4 = 0 (Posted on 2009-10-15) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Solution - trying formatting | Comment 10 of 14 |

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 + (c)*y + (c)*z = 0

Rearrange and square (1)
(2) x = (c)*y + 2c*yz +c*c*z

Equation 2 multiplied by z and rearranged
(3) x*z = ((c)*z)*y + 2c*yz +c*c*z

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

Rearrange (4) according to terms with cbrt(c)
(5) x*z + y*z - 2*c*y*z = c*(c*z - y)

The left side (x*z + y*z - 2*c*y*z) is rational.  The right factor (c*z - y) is rational, but c 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
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