In spite of the spate of unanswered questions, I would like to add one more question to the forum:
Find all n such that 2n divides 3n+1
(Hope this question too does not end up as a statistic)
edited
90 ,1 are only solutions
i suppose n is whole nos ?
3n +1 = (2+1)n +1
=8β + n(n-1).2 + 2n + 2 = 8β + 2(n2+1) = 2nα
trivially 0,1 satisfy and 2 doesnt
take n>2
2n-1α = 4β + n2 +1 implys n as odd
n= 2λ+1
4λα = 4(β +λ +λ2 ) +2
so contradicition
341no n was natural :D
Nice solution.
I came up with: For even 3n≡1 mod 4, hence 3n+1 is of the form 4k+2 i.e divisible at most by 2
. Now the sequence un = 3n+1 satisifes un+2 = 9un-8
So, for odd n>3, if there is such an n with 2n|un then, un-2 will be divisible by 8 and so on till u1 is divisible by 8 which is a contradiction as u1=4.
11I have a soln, but it was quite long.... Solved this in www.goiit.com
http://www.goiit.com/posts/list/algebra-gems-of-algebra-4-929727.htm