Larson 4.1.10

1. Jan 7, 2008

ehrenfest

[SOLVED] Larson 4.1.10

1. The problem statement, all variables and given/known data
Determine all triplets of integers (x,y,x) satisfying the equation

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

2. Relevant equations

3. 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...

2. Jan 7, 2008

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

i will work it out some other time

3. Jan 7, 2008

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.

4. Jan 7, 2008

morphism

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

I have one comment on your attempted solution:
That doesn't look correct to me. Shouldn't the right side be simply 0?

5. Jan 7, 2008

ehrenfest

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

6. Jan 7, 2008

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).

7. Jan 7, 2008

ehrenfest

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.

Last edited: Jan 7, 2008