Problem 34

**ThePerfectHacker** · August 12th 2007, 08:14 PM

Given $x,y,z\in \mathbb{Z}$ solve the Diophantine equation:
$x^3+y^3+z^3 = (x+y+z)^3$ .

**mathisfun1** · August 13th 2007, 04:18 AM

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, $(j, k, -k)$

**albi** · August 24th 2007, 01:22 AM

Is it possible to propose the next 'problem of the week'?

**mathisfun1** · August 24th 2007, 03:57 AM

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.

**ThePerfectHacker** · August 24th 2007, 08:03 AM

Originally Posted by albi

Is it possible to propose the next 'problem of the week'?

I try to get one by Sunday or Monday.

Apparently I can't post new threads, so I'll just append a problem here.

Blocked priveleges.

**albi** · August 24th 2007, 09:56 AM

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´.

**Jameson** · August 29th 2007, 08:53 PM

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!

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