# Thread: Equation with three variables

1. ## Equation with three variables

$x^3 + y^3 + z^3 = 2008$
Find all triples (x,y,z) for this equation where x, y, and z are natural numbers

I figured out all the triples just through trial but I don't understand how you're supposed to figure it out. Can someone help me?

2. ## This is

a third degree Diophantine equation...look here

Diophantine Equation--3rd Powers -- from Wolfram MathWorld....

3. Originally Posted by Dslycixdued
$x^3 + y^3 + z^3 = 2008$
Find all triples (x,y,z) for this equation where x, y, and z are natural numbers

I figured out all the triples just through trial but I don't understand how you're supposed to figure it out. Can someone help me?
The best way I see to do this is by restricting the possible cases. Say that $x\geq y\geq z$. Then we can see that $z\leq 8$ and $9 \leq x\leq 12$. This gives us the some cases to try $(x,y,z)$. And any permutation will be a solution of that.

4. Originally Posted by ThePerfectHacker
Say that $x\geq y\geq z$. Then we can see that $z\leq 8$ and $9 \leq x\leq 12$.
I sort of understand that but could you explain why $z\leq 8$ and $9 \leq x\leq 12$ ?

5. Originally Posted by Dslycixdued
I sort of understand that but could you explain why $z\leq 8$ and $9 \leq x\leq 12$ ?
Because otherwise it would be impossible to get 2008. Note, $z$ is the smallest this means $x^3+y^3+z^3\geq z^3+z^3+z^3 =3z^3$ but if $z\geq 9$ then $3z^3$ would exceede 2008. Similarly with $9\leq x$. The last constraint $x\leq 12$ comes from because if $x\geq 13$ then $x^3$ would already exceede 2008.

6. Hello,

x>y>z

We know that $x^3+y^3+z^3=2008$

$2008=x^3+y^3+z^3>z^3+z^3+z^3=3z^3$

-> $z^3<2008/3$

(2008/3.0)^(1/3) ~ 8.7

Thus z< or =8 as z is an integer.

This goes the same way for x, but by superior values.