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#1 2010-01-29 13:44:30
- majdouline
- Guest
Olym
Hii !
This Exercice is very difficult to find for me ! Can You help me :
Prove that for all integer p higher or equal to 3 it exist p of integers natural different two to two
Such as :please I want this solution for tomorow !
#2 2010-01-30 03:18:13
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: Olym
Find a way to do it when p=3, then find a way to convert a p=k solution into a p=k+1 solution.
Post again if you need more hints.
Why did the vector cross the road?
It wanted to be normal.
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