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Chukchi
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Posts: 915
Joined: December 15th, 2001, 3:43 am

2013/2014

December 23rd, 2013, 2:13 am

Let r = (2014^2013 - 1)/2013Prove that 2014^r - 1 is divisible by r^2
 
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alghosh
Posts: 60
Joined: April 28th, 2010, 9:11 am

2013/2014

December 31st, 2013, 11:34 am

1st show : 2013 divides r,=> 2014^2013-1=(2013)^2*kand r=(2013)*kNow rewrite , 2014^r-1 =2014^(2013*k)-1=(2014^(2013)-1)*[ (2014)^2013*(k-1) + ... + (2014)^2013+1]=k*((2013)^2)*[ (k*(2013)^2+1) ^ (k-1)+...+(k*(2013)^2+1)+1 ]RHS above is divisible by k^2 as well as (2013)^2 [Since k divides (k*(2013)^2+1) ^ (k-1)+...+(k*(2013)^2+1)+1 ]Sorry for this script ... Hopefully you can understand...
Last edited by alghosh on December 30th, 2013, 11:00 pm, edited 1 time in total.
 
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wileysw
Posts: 593
Joined: December 9th, 2006, 6:13 pm

2013/2014

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.
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