Given $\displaystyle x,y,z\in \mathbb{Z}$ solve the Diophantine equation:

$\displaystyle x^3+y^3+z^3 = (x+y+z)^3$.

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- Aug 12th 2007, 08:14 PMThePerfectHackerProblem 34
Given $\displaystyle x,y,z\in \mathbb{Z}$ solve the Diophantine equation:

$\displaystyle x^3+y^3+z^3 = (x+y+z)^3$. - Aug 13th 2007, 04:18 AMmathisfun1
All permutations of $\displaystyle (k, -k, j)$ for integers k and j are solutions. To see that these are the only solutions, rewrite as $\displaystyle y^3+z^3 = (x+y+z)^3 - x^3$ which, assuming $\displaystyle y \ne -z$, reduces to $\displaystyle x^2+(y+z)x+yz = 0 \Rightarrow x = -y$ or $\displaystyle x = -z$ with the other variable arbitrary, giving the solutions $\displaystyle (k, -k, j), \ (k, j, -k)$ and, by symmetry in the original equation, $\displaystyle (j, k, -k)$

- Aug 24th 2007, 01:22 AMalbi
Is it possible to propose the next 'problem of the week'?

- Aug 24th 2007, 03:57 AMmathisfun1
Apparently I can't post new threads, so I'll just append a problem here.

Source: The Art of Problem Solving (Vol 2)

Show that for any two positive numbers, RMS-AM >= GM-HM where RMS denotes the root-mean square, AM the arithmetic mean, GM the geometric mean, and HM the harmonic mean. - Aug 24th 2007, 08:03 AMThePerfectHacker
- Aug 24th 2007, 09:56 AMalbi
- Aug 29th 2007, 08:53 PMJameson