June 25th, 2017, 04:43 PM  #1 
Newbie Joined: Apr 2017 From: Canada Posts: 28 Thanks: 1  Prove the following:
Prove or provide a counterexample to the following conjecture: Assuming "p" is an odd integer, "2^p  1" will never be divisible by 3 (The MATH function wasn't working for me) Last edited by Antoniomathgini; June 25th, 2017 at 04:48 PM. 
June 25th, 2017, 08:37 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,422 Thanks: 484 Math Focus: Yet to find out.  
June 25th, 2017, 09:09 PM  #3  
Senior Member Joined: Aug 2012 Posts: 1,641 Thanks: 415  Quote:
Similarly any even power of $2$ is $1$ mod $3$ and one less than that is divisible by $3$. For example $2^4 = 16$ which is $1$ more than a multiple of $3$. Last edited by Maschke; June 25th, 2017 at 09:19 PM.  
June 25th, 2017, 11:53 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,166 Thanks: 1424 
As 3 = 4  1 divides 4$^{(p  1)/2}$  1 by the factor theorem, 2$^p$  1 = 2(4$^{(p1)/2}$  1) + 1 gives a remainder of 1 when divided by 3. 
June 26th, 2017, 07:19 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond 
Either $2^n1$ or $2^n+1$ is a multiple of $3$. [Clearly, $2^n$ is not a multiple of $3$. This implies that $2^n$ is either one less or one more than a multiple of $3$.] $$(2^n1)(2^n+1)=2^{2n}1$$ $$3x=2^{2n}1$$ $$6x=2\cdot2^{2n}2$$ $$6x+1=2^{2n+1}1$$ $$QED$$ Last edited by greg1313; July 3rd, 2017 at 01:14 PM. 

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