Why is the following G.C.D equal to $1$: $$ \gcd(3^s, 2^n-3^{(j-i)}2^m),\quad s> j >i \geq 0, $$ and all variables are natural numbers.
Why the following G.C.D is $1$
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elementary-number-theory
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0If $i$ ande $j$ are switched it looks right, and easy to show. – 2012-11-21
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0@tlh1987 Because $i>j \implies i-j>0 \implies j-i<0 \implies 3^{j-i} \not\in \mathbb{N}$ – 2012-11-21
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0@joriki, sorry, there something wrong with the material I have, can I change the question? I am first asked here? – 2012-11-21
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0@badp sorry, I swithed the i and j. – 2012-11-21
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0@tlh1987: When you edit the question in a way that makes an existing answer or comment appear wrong, please mark the edit as such. Thanks. – 2012-11-21
2 Answers
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The only prime factor of $3^s$ is 3 as $s\ge 1$
But $2^n-3^{(j-i)}2^m\equiv2^n\pmod 3$ as $3\mid 3^{j-i}$ as $j>i$
So, $2^n-3^{(j-i)}2^m\equiv2^n\equiv(-1)^n\not\equiv 0 \pmod 3$
So, $3^s,2^n-3^{(j-i)}2^m$ can not have any common prime factor, hence $(3^s,2^n-3^{(j-i)}2^m)=1$
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0sorry, I switched the i and j, is this still the answer? – 2012-11-21
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0@tlh1987, please find the edited answer. – 2012-11-21
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0Excuse my ignorance... what does the bar in $j>i,3\mid 3^{j-i}$ mean? I thought it meant "such that," but I can't parse it in a way that makes sense with a 3 before it. – 2012-11-21
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1@LarsH, $a\mid b$ means $b$ is divisible by $a$. Nothing wrong in clarifying doubt. – 2012-11-21
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0Thanks, that clears it up. I also had to look up the notation $(2,3) = 1$. Apparently that's short for $ \gcd(2,3) = 1 $. – 2012-11-21
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0@LarsH, may have a look in to the Notations in http://books.google.co.in/books?id=V52HIcKguJ4C&printsec=frontcover&dq=inauthor:%22Herbert+S.+Zuckerman%22&hl=en&sa=X&ei=CAitUPG_JoizrAeC_4DgDg&ved=0CDAQ6AEwAA#v=onepage&q&f=false – 2012-11-21
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Laws of GCD:
- $$\gcd(x,y) = \gcd(x,x-y)$$
- for $a$ coprime to $y$: $$\gcd(x,y) = \gcd(x,ay)$$
We can derive general formula using the laws of GCD:
$$\gcd(3^a,2^b) = 1$$
$$\gcd(3^a,2^b-3^a) = 1$$
$$\gcd(3^a,2^{b+c}-2^c 3^a) = 1$$
$$\gcd(3^{a+d},2^{b+c}-2^c 3^a) = 1$$
now put $a+d = s$, $b+c = n$, $a = j-i$, $c = m$ to get the special result.
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1thanks! Your answer is very clear! – 2012-11-21