Problem 34

ThePerfectHacker · #1 $Old$ August 12th, 2007, 10:14 PM

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

mathisfun1 · #2 $Old$ August 13th, 2007, 06:18 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 · #3 $Old$ August 24th, 2007, 03:22 AM

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

mathisfun1 · #4 $Old$ August 24th, 2007, 05:57 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.