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 June 25th, 2017, 04:43 PM #1 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 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
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Quote:
 Originally Posted by Antoniomathgini Assuming $p$ is an odd integer, $2^p - 1$ will never be divisible by $3$ (The MATH function wasn't working for me)
What do you mean it wasn't working?

June 25th, 2017, 09:09 PM   #3
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Quote:
 Originally Posted by Antoniomathgini 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)
$2 \equiv -1 \pmod 3$ so any odd power of $2$ must be $-1$ mod $3$. Subtracting one gives a result that's $-2$ mod $3$.

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: 18,594 Thanks: 1492 As 3 = 4 - 1 divides 4$^{(p - 1)/2}$ - 1 by the factor theorem, 2$^p$ - 1 = 2(4$^{(p-1)/2}$ - 1) + 1 gives a remainder of 1 when divided by 3. Thanks from topsquark and Antoniomathgini
 June 26th, 2017, 07:19 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,743 Thanks: 1001 Math Focus: Elementary mathematics and beyond Either $2^n-1$ 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 $3$.] $$(2^n-1)(2^n+1)=2^{2n}-1$$ $$3x=2^{2n}-1$$ $$6x=2\cdot2^{2n}-2$$ $$6x+1=2^{2n+1}-1$$ $$QED$$ Thanks from topsquark and Antoniomathgini Last edited by greg1313; July 3rd, 2017 at 01:14 PM.

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