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 October 11th, 2017, 06:11 PM #1 Newbie   Joined: Oct 2017 From: Dream Country Posts: 3 Thanks: 1 Help for the hard prove question If n is a positive integer and n is even, prove: (2^(n!)-1) is divisible by (n^2-1). This question confuse me some days. Please help me or give me some hint. Thank you very much. Last edited by skipjack; October 12th, 2017 at 01:57 AM. October 11th, 2017, 07:27 PM   #2
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 Originally Posted by GGQ If n is a positive integer and n is even, prove: (2^(n!)-1) is divisible by (n^2-1). This question confuse me some days. Please help me or give me some hint. Thank you very much.
Let n = 3. Then $\displaystyle \frac{2^{3!} - 1}{3^2 - 1} = \frac{2^6 - 1}{8}$ $\displaystyle = \frac{63}{8}$
So it doesn't work for n = 3.

-Dan

Last edited by skipjack; October 12th, 2017 at 01:58 AM. October 11th, 2017, 07:44 PM #3 Newbie   Joined: Oct 2017 From: Dream Country Posts: 3 Thanks: 1 question is " n is even " so the example that you take n=3 is illegal. thank you. Thanks from topsquark October 12th, 2017, 11:35 AM   #4
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 Originally Posted by GGQ question is " n is even " so the example that you take n=3 is illegal. thank you.
Sorry. Yes, I missed that.

-Dan October 12th, 2017, 12:29 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for "proof by induction" on k. Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes "2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3. Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1. Thanks from topsquark and GGQ Last edited by Country Boy; October 12th, 2017 at 12:33 PM. October 12th, 2017, 05:58 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Don't you need it to be divisible by $(2k)^2-1=4k^2-1$? Thanks from topsquark October 12th, 2017, 10:09 PM   #7
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Quote:
 Originally Posted by Country Boy Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for "proof by induction" on k. Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes "2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3. Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1.

This part " 2^((2k)!-1) is divisible by (2k)^2- 1. " is wrong.
Correct is ( 2^(2k)!) -1 is divisible by (2k)^2- 1
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