
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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^21). 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: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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. 
October 12th, 2017, 11:35 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  
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+11)= 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. 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: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Don't you need it to be divisible by $(2k)^21=4k^21$?

October 12th, 2017, 10:09 PM  #7  
Newbie Joined: Oct 2017 From: Dream Country Posts: 3 Thanks: 1  Quote:
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  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A really hard math problem "Prove that if..."  isel  Algebra  8  September 22nd, 2013 09:35 AM 
hard Trigonometry prove  stuart clark  Algebra  1  March 23rd, 2011 11:46 PM 
very hard question  riotsandravess  Advanced Statistics  4  November 3rd, 2010 06:22 PM 
very hard question  riotsandravess  Advanced Statistics  0  October 28th, 2010 01:21 PM 
Hard prove !  yehoram  Algebra  8  August 9th, 2010 08:37 PM 