# Thread: integer triples

1. ## integer triples

i am trying to find all the integer triples that satisfy
x^3 +y^3 +z^3 =(x+y+z)^3

i got as far as x^3+y^3+z^3=x^3+3x^2y+3y^2z+3x^2z+6xyz+3xy^2+3xz^2 +y^3+3yz^2+z^3
=3x^2y+3y^2z+3x^2z+6xyz+3xy^2+3xz^2+3yz^2
i know that all of the terms are divisible by 3 but i am not sure exactly how to proceed.

2. Originally Posted by cookiecutter
i am trying to find all the integer triples that satisfy
x^3 +y^3 +z^3 =(x+y+z)^3

i got as far as x^3+y^3+z^3=x^3+3x^2y+3y^2z+3x^2z+6xyz+3xy^2+3xz^2 +y^3+3yz^2+z^3
=3x^2y+3y^2z+3x^2z+6xyz+3xy^2+3xz^2+3yz^2
i know that all of the terms are divisible by 3 but i am not sure exactly how to proceed.
What you have then is that
$\displaystyle 3x^{2}y+3y^{2}z+3x^{2}z+6xyz+3xy^{2}+3xz^{2}+3yz^{ 2}=0$
since all the cubed terms cancel from either side of your original equation.

Dividing by 3 will simplify this a little to
$\displaystyle x^{2}y+y^{2}z+x^{2}z+2xyz+xy^{2}+xz^{2}+yz^{2}=0$

Now the trivial solution is obvious $\displaystyle (x,y,z)=(0,0,0)$

The next step I would advise is setting one of the variables, let's say $\displaystyle x=0$ then we have
$\displaystyle y^{2}z+yz^{2}=0 \implies y+z=0 \implies y=-z$
so we have a solution of the form $\displaystyle (x,y,z)=(0,\alpha,-\alpha)$
where $\displaystyle \alpha$ is any non zero integer (otherwise it's a repeat of the trivial solution)

You can then use the same argument for $\displaystyle y=0, z=0$ but the solutions should be very similar (and hopefully obvious).

Finally assume all are non zero.
If all are positive then every term in the sum is positive and thus cannot be zero, similarly if all are negative. So we can assume that $\displaystyle x$ is negative and the other two are positive. If we let $\displaystyle x=-\beta$ where $\displaystyle \beta$ is any positive integer we get
$\displaystyle \beta^{2}y+y^{2}z+\beta^{2}z+yz^{2}=2\beta yz + \beta y^{2} + \beta z^{2}$
which simplifies to
$\displaystyle (\beta^{2}+yz)(y+z)=\beta(y+z)^{2}$
which simplifies further to
$\displaystyle \beta^{2}+yz=\beta(y+z)$

Applying the quadratic formula to this will yield the solution $\displaystyle \beta=y,z$

So we have another solution set $\displaystyle (x,y,z)=(-\beta,\beta,\beta)$ and again the other two variables will yield similar solutions.

As far as I can tell this forms a complete set of solutions.
Hope this helps.