I'm stuck. Solving Larson's Equation: 4.1.10

[ehrenfest]

[SOLVED] Larson 4.1.10

Homework Statement

Determine all triplets of integers (x,y,x) satisfying the equation

x^3+y^3+z^3=(x+y+z)^3

Homework Equations

The Attempt at a Solution

I think the only solutions are x=y=0, x=z=0, z=y=0, z=-y and x=0, x=-z and y=0, x=-y and z=0. If I rewrite the equation as (x+y)(x+z)(y+z)=-xyz, several things are obvious:
1)x,y,z are not all odd
2)x,y,z are not all positive and not all negative
3)if for example, z is negative and x,y are positive, then abs(z)< x,y or abs(z)>x,y

But how can I show there are no other solutions...

 

[sadhu]

certainly either all are 0 or none
by fermat last theorem(diff. from earliar one)

i will work it out some other time

 

[ehrenfest]

I am sure there is a way to do this without using Fermat's Last Theorem. None of Larson's problems require something that advanced.

 

[morphism]

I don't even see how Fermat is relevant here.

I have one comment on your attempted solution:

If I rewrite the equation as (x+y)(x+z)(y+z)=-xyz

That doesn't look correct to me. Shouldn't the right side be simply 0?

 

[ehrenfest]

That doesn't look correct to me. Shouldn't the right side be simply 0?

No. Multiply it out. The LHS only gives 2xyz.

 

[morphism]

Maybe it's the sleepiness, but it seems to me that (x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(x+z)(y+z).

 

[ehrenfest]

Maybe it's the sleepiness, but it seems to me that (x+y+z)^3 = x^3 + y^3 + z^3 + 3(x+y)(x+z)(y+z).

You're right. It does. The solutions to (x+y)(x+z)(y+z)=0 (and to the the original equation) are x=-y,y=-z, or z=-x.