# Find all integer numbers

• May 29th 2010, 11:05 AM
dhiab
Find all integer numbers
Find all integer numbers (x,y,z) :
$
x^{3}+y^{3}+z^{3}=2008
$
• May 29th 2010, 12:32 PM
oldguynewstudent
Quote:

Originally Posted by dhiab
Find all integer numbers (x,y,z) :
$
x^{3}+y^{3}+z^{3}=2008
$

The only integers that I've found to work are (12,6,4) and (10,10,2).

So if you assign these ordered triples to x,y, and z and cycle the values you should get 12 different integer combinations of x,y, and z that will solve the equations.
• May 29th 2010, 02:44 PM
chiph588@
Quote:

Originally Posted by dhiab
Find all integer numbers (x,y,z) :
$
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 $x,y,z<\sqrt[3]{2008}\approx12.615987<13$ 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 $9$ solutions.
However if you disregard order there are $2$ solutions namely $(x,y,z)=(12,6,4) \text{ and } (10,10,2)$, as oldguynewstudent has already pointed out.

Edit: I just realized you wanted your equation solved over $\mathbb{Z}$, not $\mathbb{N}$. This throws my idea right out the window...

May I ask what this is for?
• May 29th 2010, 03:28 PM
Xitami
Code:

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
surely more
• May 29th 2010, 05:37 PM
chiph588@
Quote:

Originally Posted by dhiab
Find all integer numbers (x,y,z) :
$
x^{3}+y^{3}+z^{3}=2008
$

Not sure all solutions are known by anyone...

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