## ganeshie8 one year ago show that $\large \dfrac{3^n-1}{2^n-1}$ is never an integer for $$n\gt 1$$ and $$n\in\mathbb{Z}$$

1. anonymous

can u make it clear pls

2. ganeshie8

3. anonymous

still not clear

4. ganeshie8

basically the problem is about showing that expression always evaluates to some non integer

5. anonymous

i should know the statement rite?

6. anonymous

*

7. dan815

even/odd?

8. dan815

thats a start

9. ganeshie8

looks like a good start! i know $$3\mid (2^n-1)$$ when $$n$$ is even

10. dan815

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11. anonymous

It will be a rational number if $2^n-1\neq0$ 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

12. dan815

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13. dan815

ok i see why 3 divides it now

14. ganeshie8

woah you always find a way to use binomial theorem !

15. ikram002p

when n is even :- 3|2^n-1 thus 2^n-1=3k bur 3 do not divide 3^n-1 3^n-1 cant be of the form 3f

16. dan815

lols its all i know

17. anonymous

ganeshie can u check my post?see above

18. ganeshie8

im pretty sure you can derive below identity with binomial thm too $4^n-1^n = (4-1)(4^{n-1}+4^{n-2}+\cdots+1) = 3(stuff)$

19. anonymous

no wait it seems wrong XD

20. ganeshie8

@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

21. dan815

i think this means we solved it for even atleast

22. ikram002p

huh duh

23. dan815

we dont need u here U OS HATER

24. anonymous

hmmmm

25. ganeshie8

haha when ppl say i hate myself, they don't literally mean it

26. ikram002p

in general k^n-1 do not divide g^n-1 for gcd(k,g)=1 both primes but that need primitive thingy

27. ikram002p

no no no dan i dont hate os :O

28. dan815

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29. ikram002p

ok i have other idea

30. ganeshie8

so are you saying $\large \dfrac{g^n-1}{k^n-1}$ is never an integer when $$\gcd(g,k)=1$$ ?

31. ganeshie8

@ikram002p

32. ikram002p

yes

33. ikram002p

well no -.-

34. dan815

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35. ikram002p

only when k is 2 (sorry not focused )

36. ganeshie8

okay odd case looks a bit more involved

37. dan815

the bottom is 2 mod 3

38. dan815

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39. ganeshie8

top is 2 mod 3 bottom is 1 mod 3

40. ikram002p

cool

41. ganeshie8

i didn't get how that helps

42. dan815

oh nvm thats just silly

43. dan815

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44. dan815

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45. dan815

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46. dan815

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47. dan815

does any of this help

48. ikram002p

can we try mod 5 -.-

49. dan815

or 7

50. dan815

can we do induction maybe

51. UsukiDoll

@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

52. dan815

what does fermats little theorem say again

53. UsukiDoll

If p is a prime number then for any integer a , the number $a^p-a$ is an integer multiple of p. IN the notation of modular arithmetic. this is expressed as $a^p \equiv a (\mod p)$

54. ikram002p

55. UsukiDoll

If a is not divisible by p, Fermat's Little Theorem is equivalent to the statement that $a^{p-1} -1$ is an integer multiple of p or in symbols $a^{p-1} \equiv 1 (\mod p)$

56. ikram002p

bbye every one :)

57. UsukiDoll

$\large \dfrac{3^n-1}{2^n-1}$ looks similar to the a not divisible by p . I feel like letting n =p-1 but that's just silly

58. dan815

did u find a solution to this yetq

59. UsukiDoll

no. T_T

60. ganeshie8

me neither, but i feel there exists a simple way for the odd case too

61. UsukiDoll

who came up with this T_T