# Thread: Find all integer numbers

1. ## Find all integer numbers

Find all integer numbers (x,y,z) :
$\displaystyle x^{3}+y^{3}+z^{3}=2008$

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

3. 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 $\displaystyle x,y,z<\sqrt{2008}\approx12.615987<13$ 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?

4. 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

5. Originally Posted by dhiab Find all integer numbers (x,y,z) :
$\displaystyle x^{3}+y^{3}+z^{3}=2008$
Not sure all solutions are known by anyone...

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

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