Find all integer numbers (x,y,z) :

$\displaystyle

x^{3}+y^{3}+z^{3}=2008

$

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- May 29th 2010, 11:05 AMdhiabFind all integer numbers
**Find all integer numbers (x,y,z) :**

$\displaystyle

x^{3}+y^{3}+z^{3}=2008

$ - May 29th 2010, 12:32 PMoldguynewstudent
- May 29th 2010, 02:44 PMchiph588@
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;

}}}}

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)

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? - May 29th 2010, 03:28 PMXitamiCode:
`2550 -2466 -1166`

1735 -1654 -887

1222 -1058 -862

361 -289 -284

13 -5 -4

-6 13 3

6 12 4

2 10 10

-113 108 57

-180 166 108

-1340 1336 278

-1152 1128 454

-5274 5269 747

-3770 3760 752

-2276 2246 772

- May 29th 2010, 05:37 PMchiph588@
Not sure all solutions are known by anyone...

http://www.mathematik.uni-bielefeld....S/elkies_cubic