# Divisibility

• Sep 21st 2011, 08:15 AM
gollum
Divisibility
I am so sorry for misleading you. I needed to solve this problem, because it occurred to me whilst I was solving another one. Indeed, what I had written is not true. In such case, could you help me find all integers $\displaystyle n >= 1$ for which
$\displaystyle 1+2^{n+1} + 2^{2n+2}$

i divisible by

$\displaystyle 1+2^{n} + 2^{2n}$?
• Sep 21st 2011, 10:18 AM
Opalg
Re: Divisibility
Quote:

Originally Posted by gollum
Could somebody help me prove that for any integer $\displaystyle n >= 1$
$\displaystyle 1+2^{n+1} + 2^{2n+2}$

i divisible by

$\displaystyle 1+2^{n} + 2^{2n}$?

I have already tried proving it using induction but I got stuck in the final step where n=k+1. Please, help.

Have you checked what happens if you try putting n=2, or n=3?
• Sep 21st 2011, 10:25 AM
HallsofIvy
Re: Divisibility
It is impossible to prove something that is NOT true! What reason do you have to believe that this statement is true?
• Sep 21st 2011, 11:25 AM
gollum
Re: Divisibility
Hi! I have already corrected the post.
• Sep 21st 2011, 12:03 PM
Opalg
Re: Divisibility
You could try to show that the ratio $\displaystyle \frac{1+2^{n+1} + 2^{2n+2}}{1+2^{n} + 2^{2n}}$ increases towards a limiting value 4. The ratio is equal to 3 when n=1. For n>1, it will lie between 3 and 4, so can never again be an integer.
• Sep 21st 2011, 12:33 PM
gollum
Re: Divisibility
Thanks, but how do I find maximum and minimum of a function like this - I mean it is exponential as well as rational.
• Sep 21st 2011, 11:40 PM
Opalg
Re: Divisibility
Probably the easiest approach is to write $\displaystyle 4 -\frac{1+2^{n+1} + 2^{2n+2}}{1+2^{n} + 2^{2n}}$ in the form $\displaystyle \frac{???}{1+2^{n} + 2^{2n}}$. Then show that that fraction is positive and less than 1 when n>1.