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April 22nd, 2014, 02:52 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, 05:11 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  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, 05:30 AM  #3 
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  
April 22nd, 2014, 06:01 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, 06:06 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, 06:22 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, 06:37 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. 

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