My Math Forum Help for the hard prove question

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 1,879
Thanks: 761

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 1,879
Thanks: 761

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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: 3,261 Thanks: 894 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: 7,355 Thanks: 2469 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
Newbie

Joined: Oct 2017
From: Dream Country

Posts: 3
Thanks: 1

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
thank you.

 Tags hard, integer, prove, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post isel Algebra 8 September 22nd, 2013 09:35 AM stuart clark Algebra 1 March 23rd, 2011 11:46 PM riotsandravess Advanced Statistics 4 November 3rd, 2010 06:22 PM riotsandravess Advanced Statistics 0 October 28th, 2010 01:21 PM yehoram Algebra 8 August 9th, 2010 08:37 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top