# Problem 34

• Aug 12th 2007, 09:14 PM
ThePerfectHacker
Problem 34
Given $x,y,z\in \mathbb{Z}$ solve the Diophantine equation:
$x^3+y^3+z^3 = (x+y+z)^3$.
• Aug 13th 2007, 05:18 AM
mathisfun1
All permutations of $(k, -k, j)$ for integers k and j are solutions. To see that these are the only solutions, rewrite as $y^3+z^3 = (x+y+z)^3 - x^3$ which, assuming $y \ne -z$, reduces to $x^2+(y+z)x+yz = 0 \Rightarrow x = -y$ or $x = -z$ with the other variable arbitrary, giving the solutions $(k, -k, j), \ (k, j, -k)$ and, by symmetry in the original equation, $(j, k, -k)$
• Aug 24th 2007, 02:22 AM
albi
Is it possible to propose the next 'problem of the week'?
• Aug 24th 2007, 04:57 AM
mathisfun1
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, 09:03 AM
ThePerfectHacker
Quote:

Originally Posted by albi
Is it possible to propose the next 'problem of the week'?

I try to get one by Sunday or Monday.

Quote:

Apparently I can't post new threads, so I'll just append a problem here.
Blocked priveleges.
• Aug 24th 2007, 10:56 AM
albi
Quote:

Originally Posted by ThePerfectHacker
I try to get one by Sunday or Monday.

Um. I think we didnt understand ourselves.

I meant I have one idea which may become ´problem of the week´.
• Aug 29th 2007, 09:53 PM
Jameson
Quote:

Originally Posted by albi
Um. I think we didnt understand ourselves.

I meant I have one idea which may become ´problem of the week´.

Of course you can! We're always open to suggestions. The Problem of the Week writers (CB and TPH) work very hard though so be constructive! :)