Problem !! Problem !!!

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sm.joty
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Problem !! Problem !!!

Unread post by sm.joty » Sun Apr 01, 2012 2:33 pm

If $x,y,z,n \in\mathbb{N}$, and $n\geq z$ then the relation does not hold $x^n+y^n= z^n$ ;)
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*Mahi*
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Re: Problem !! Problem !!!

Unread post by *Mahi* » Sun Apr 01, 2012 3:33 pm

sm.joty wrote:If $x,y,z,n \in\mathbb{N}$, and $n\geq z$ then the relation does not hold $x^n+y^n= z^n$ ;)
Without FLT please ;)
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Phlembac Adib Hasan
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Re: Problem !! Problem !!!

Unread post by Phlembac Adib Hasan » Sun Apr 01, 2012 4:44 pm

Solution:
WLOG we may assume $x\ge y$.Surely if $y=1$, it has no solution.
$z\ge x+1$.So $n\ge x+1$.
Prove $(x+1)^n\ge 2x^n$ for $x\ge 2$ and $n\ge x+1$
Therefore $(x+1)^n\ge 2x^n\ge x^n+y^n=z^n$.
So it has no solution. :)
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zadid xcalibured
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Re: Problem !! Problem !!!

Unread post by zadid xcalibured » Sun Apr 01, 2012 5:16 pm

this problem has nothing to do with number theory.total inequality.

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