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April 22nd, 2014, 02:52 AM   #1
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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 ?
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April 22nd, 2014, 05:11 AM   #2
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Quote:
Originally Posted by gelatine1 View Post
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.
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April 22nd, 2014, 05:30 AM   #3
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Originally Posted by Olinguito View Post
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?
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April 22nd, 2014, 06:01 AM   #4
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I’m referring to $3a\equiv3b\pmod{27}\,\Rightarrow\,a\equiv b\pmod{27}$.
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April 22nd, 2014, 06:06 AM   #5
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Oh, I see what you mean. Yes, that works as long as 3 is coprime with ab, I agree.
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April 22nd, 2014, 06:22 AM   #6
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Actually I made a mistake. It won’t work.

$6\equiv33\pmod{27}$ but $2\not\equiv11\pmod{27}$.
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April 22nd, 2014, 06:37 AM   #7
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Oops. Yes, I guess you really need to do it mod 27/3.
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