Originally Posted by
dhiab Find all integer numbers (x,y,z) :
$\displaystyle
x^{3}+y^{3}+z^{3}=2008
$
Sometimes I prefer brute force over deriving the solution in a creative way... This is one of those times.
We must have $\displaystyle x,y,z<\sqrt[3]{2008}\approx12.615987<13 $ so there are only finitely many cases to consider.
I wrote code to test these cases. Now it's not the most efficient code but hey it gets the job done.
Code:
for(int x=0;x<13;x++){
for(int y=0;y<13;y++){
for(int z=0;z<13;z++){
if(x*x*x+y*y*y+z*z*z==2008){
std::cout<<"(x,y,z)=("<<x<<","<<y<<","<<z<<")"<<std::endl;
}}}}
And here is the output i.e. all solutions:
Code:
(x,y,z)=(2,10,10)
(x,y,z)=(4,6,12)
(x,y,z)=(4,12,6)
(x,y,z)=(6,4,12)
(x,y,z)=(6,12,4)
(x,y,z)=(10,2,10)
(x,y,z)=(10,10,2)
(x,y,z)=(12,4,6)
(x,y,z)=(12,6,4)
So as you can see there are $\displaystyle 9 $ solutions.
However if you disregard order there are $\displaystyle 2 $ solutions namely $\displaystyle (x,y,z)=(12,6,4) \text{ and } (10,10,2) $, as oldguynewstudent has already pointed out.
Edit: I just realized you wanted your equation solved over $\displaystyle \mathbb{Z} $, not $\displaystyle \mathbb{N} $. This throws my idea right out the window...
May I ask what this is for?