Unrelentingly Yours

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

4 Answers

9
Celestine preetham ·

is it 3n+1 or something else ?

9
Celestine preetham ·

0 ,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

341
Hari Shankar ·

no 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.

11
Devil ·

I 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

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