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[SOLVED] Divisibility challenge

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,802
Let $x,\,y,\,z$ be integers such that $(x-y)^2+(y-z)^2+(z-x)^2=xyz$, prove that $x^3+y^3+z^3$ is divisible by $x+y+z+6$.
 

kaliprasad

Well-known member
Mar 31, 2013
1,331
Let $x,\,y,\,z$ be integers such that $(x-y)^2+(y-z)^2+(z-x)^2=xyz$, prove that $x^3+y^3+z^3$ is divisible by $x+y+z+6$.
We know $x^3+y^3+z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$

Hence $x^3+y^3+z^3 = 3xyz + \frac{1}{2}(x+y+z)(xyz)$ (putting the value from given condition)

Or $x^3+y^3+z^3 = xyz( 3 + \frac{1}{2}(x+y+z))$

Or $x^3+y^3+z^3 = \frac{xyz}{2}( 6 + x+y+z)$

If we can prove that xyz is even then we are through

As (x-y), (y-z) and (z-x) sum to give zero so atleast one of them is even. So xyz is even from the given condition so $\frac{xyz}{2}$ is an integer and hence $x^3+y^3+z^3$ is multiple of $(6 + x+y+z)$