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Topic:
for all integers x coprime to n, x^(n1) = 1 mod n > that n square free
Replies:
2
Last Post:
Oct 25, 2009 6:48 PM




Re: for all integers x coprime to n, x^(n1) = 1 mod n > that n square free
Posted:
Oct 25, 2009 6:48 PM


> G Patel <gaya.patel@gmail.com> writes: > > > How can I show that if for all integers x coprime to n, x^(n1) = 1 > > (mod n), then n is necessarily square free? > > Hint: Your hypothesis implies phi(n) divides n1. > But if p^2 divides n for some prime p, then p divides phi(n).
Oops, sorry, both of those phi's should be lambda's (lambda(n) = Carmichael function).  Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada



