J jazzman New member Joined Jan 20, 2008 Messages 18 Jun 3, 2008 #1 Hey all! I need to prove that 30 is a divisor of \(\displaystyle n^5-n\) for \(\displaystyle n\ge3\). I tried expanding: \(\displaystyle n^5-n=n(n^4-1)=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1)\). But this doesn't seem to help. Please help!
Hey all! I need to prove that 30 is a divisor of \(\displaystyle n^5-n\) for \(\displaystyle n\ge3\). I tried expanding: \(\displaystyle n^5-n=n(n^4-1)=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1)\). But this doesn't seem to help. Please help!
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jun 3, 2008 #2 Re: prove [n^5-n] divides by 30 See here: http://mathforum.org/library/drmath/view/60840.html
J jazzman New member Joined Jan 20, 2008 Messages 18 Jun 3, 2008 #3 Re: prove [n^5-n] divides by 30 Thanks a lot for the link! galactus said: See here: http://mathforum.org/library/drmath/view/60840.html Click to expand...
Re: prove [n^5-n] divides by 30 Thanks a lot for the link! galactus said: See here: http://mathforum.org/library/drmath/view/60840.html Click to expand...
