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ganeshie8
 one year ago
show that \[\large \dfrac{3^n1}{2^n1}\] is never an integer for \(n\gt 1\) and \(n\in\mathbb{Z}\)
ganeshie8
 one year ago
show that \[\large \dfrac{3^n1}{2^n1}\] is never an integer for \(n\gt 1\) and \(n\in\mathbb{Z}\)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u make it clear pls

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0hey please check now @chethus

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0basically the problem is about showing that expression always evaluates to some non integer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i should know the statement rite?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0looks like a good start! i know \(3\mid (2^n1)\) when \(n\) is even

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It will be a rational number if \[2^n1\neq0\] as both numerator and denominator will be integers clearly \[2^n \neq 1\]\[n \log2 \neq \log1\]\[n \log 2 \neq 0\]\[n \neq 0\] which is satisfied

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ok i see why 3 divides it now

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0woah you always find a way to use binomial theorem !

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1when n is even : 32^n1 thus 2^n1=3k bur 3 do not divide 3^n1 3^n1 cant be of the form 3f

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ganeshie can u check my post?see above

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0im pretty sure you can derive below identity with binomial thm too \[4^n1^n = (41)(4^{n1}+4^{n2}+\cdots+1) = 3(stuff)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no wait it seems wrong XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0@Nishant_Garg yes the given expression is always a rational number for \(n\gt 1\) but that doesn't prove it can never be an integer right

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i think this means we solved it for even atleast

dan815
 one year ago
Best ResponseYou've already chosen the best response.2we dont need u here U OS HATER

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0haha when ppl say i hate myself, they don't literally mean it

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1in general k^n1 do not divide g^n1 for gcd(k,g)=1 both primes but that need primitive thingy

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1no no no dan i dont hate os :O

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ok i have other idea

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so are you saying \[\large \dfrac{g^n1}{k^n1}\] is never an integer when \(\gcd(g,k)=1\) ?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1only when k is 2 (sorry not focused )

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0okay odd case looks a bit more involved

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0top is 2 mod 3 bottom is 1 mod 3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i didn't get how that helps

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh nvm thats just silly

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1can we try mod 5 .

dan815
 one year ago
Best ResponseYou've already chosen the best response.2can we do induction maybe

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1@chethus this is a proof problem (pure math) which means we have to use different definitions or theorems to prove or disprove a statement. At first, it's hard to grasp because pure math is way different than computation math which is easily found in elementary, middle, or high schools

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what does fermats little theorem say again

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1If p is a prime number then for any integer a , the number \[a^pa\] is an integer multiple of p. IN the notation of modular arithmetic. this is expressed as \[a^p \equiv a (\mod p) \]

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i give up, have no clue to start with.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1If a is not divisible by p, Fermat's Little Theorem is equivalent to the statement that \[a^{p1} 1 \] is an integer multiple of p or in symbols \[a^{p1} \equiv 1 (\mod p)\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \dfrac{3^n1}{2^n1} \] looks similar to the a not divisible by p . I feel like letting n =p1 but that's just silly

dan815
 one year ago
Best ResponseYou've already chosen the best response.2did u find a solution to this yetq

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0me neither, but i feel there exists a simple way for the odd case too

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1who came up with this T_T
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