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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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Quote:
 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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Quote:
 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

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