January 1st, 2014, 2:08 am
one simply needs binomial expansion for this one:as algosh pointed out above, r=[(2013+1)^2013-1]/2013=[O(2013^2)+(2013 choose 1)*2013]/2013, so r is divisible by 2013, thus 2014^r-1=(2014^2013)^(r/2013)-1=(2013*r+1)^(r/2013)-1=O(r^2)+(r/2013 choose 1)*(2013*r) is divisible by r^2----- ----- ----- ----- -----in case this is not clear, but 2013 can be replaced by any integer and the divisibility would still hold
Last edited by
wileysw on January 20th, 2014, 11:00 pm, edited 1 time in total.