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 April 22nd, 2014, 03:52 AM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 modulo operations Hi I was wondering if it is allowed to do something like this: if $\displaystyle a \equiv b \pmod{9}$ then $\displaystyle 3a \equiv 3b \pmod{27}$ or $\displaystyle a \equiv b \pmod{27}$ I wanted to use this for the following problem: prove that $\displaystyle 3^{n+1} | 2^{3^n} + 1$ for every $\displaystyle n \in N$ I wanted to use induction for this problem. And I said if this expression holds for a value k, then does it hold too for a value k+1? this means that $\displaystyle 3^{k+1} | 2^{3^k} + 1$ or $\displaystyle 2^{3^k} \equiv -1 \pmod {3^{k+1}}$ If we cube the expression we get $\displaystyle 2^{3*3^k} \equiv -1 \pmod {3^{k+1}}$ and now if I would be allowed to use the identity I had said then I would say that $\displaystyle 2^{3^{k+1}} \equiv -1 \pmod {3^{k+2}}$ which means that $\displaystyle 3^{k+2} | 2^{3^{k+1}} + 1$ if we fill in the expression for zero we would conclude that this is true for every value of n. But I don't think I can use that identity I had given because that allows me to prove that $\displaystyle 3^{k+m} | 2^{3^{k+1}} + 1$ for every m which is obviously not true. Could anyone help me out ? April 22nd, 2014, 06:11 AM   #2
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 Originally Posted by gelatine1 Hi I was wondering if it is allowed to do something like this: if $\displaystyle a \equiv b \pmod{9}$ then $\displaystyle 3a \equiv 3b \pmod{27}$ or $\displaystyle a \equiv b \pmod{27}$
This is fine as long as neither $a$ nor $b$ is divisible by $3$. Therefore it is okay in your induction proof. April 22nd, 2014, 06:30 AM   #3
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 Originally Posted by Olinguito This is fine as long as neither $a$ nor $b$ is divisible by $3$. Therefore it is okay in your induction proof.
I think it's fine even then. 3 = 12 mod 9 and also 9 = 36 mod 27, right? April 22nd, 2014, 07:01 AM #4 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra I’m referring to $3a\equiv3b\pmod{27}\,\Rightarrow\,a\equiv b\pmod{27}$. April 22nd, 2014, 07:06 AM #5 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Oh, I see what you mean. Yes, that works as long as 3 is coprime with ab, I agree. April 22nd, 2014, 07:22 AM #6 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Actually I made a mistake. It won’t work. $6\equiv33\pmod{27}$ but $2\not\equiv11\pmod{27}$. April 22nd, 2014, 07:37 AM #7 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Oops. Yes, I guess you really need to do it mod 27/3. Thanks from Olinguito Tags modulo, operations Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Drake Number Theory 3 April 7th, 2013 10:41 AM zelmac Number Theory 1 December 28th, 2012 10:57 AM adam2009 Algebra 2 May 10th, 2010 01:31 PM swtrse Number Theory 2 August 6th, 2009 05:53 AM julien Abstract Algebra 1 November 19th, 2006 09:51 PM

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