Question

Challenge problem : Show that \(F(a) = \frac{2^{2^{a+1}}-1}{2^{2^a}+1}\) is an integer for all \(a \in \mathbb{N} \)

Answer 1

\[\large F(a) = \frac{2^{2^{a+1}}-1}{2^{2^a}+1} \]

Answer 2

i think i can answer it

Answer 3

@bohotness im sure you can... please go ahead :)

Answer 4

i have it on my phone and i try to frigre the rest out

Answer 5

but i guess you need to find out what a is

Answer 6

\[2^{2^1}+1=2^2+1=4+1=5 \\ 2^{2^{1+1}}-1=2^{2^{2}}-1=16-1=15 \\ \text{ and looks good for } a=1 \text{ since } 5|15\]

Answer 7

2^2(a+1)-1

Answer 8

suppose \(\large 2^{2^a}+1~ |~2^{2^{a+1}}-1 \)

we need to show \(\large 2^{2^{a+1}}+1~ |~2^{2^{a+2}}-1 \)

Answer 9

k

Answer 10

2*l+220*i+221

Answer 11

difference of squares

Answer 12

that will do haha! xD

Answer 13

\[\large F(a) = \frac{2^{2^{a+1}}-1}{2^{2^a}+1} = \frac{2^{2\cdot {2^{a}}}-1}{2^{2^a}+1} = \frac{\left(2^{ {2^{a}}}\right)^2-1}{2^{2^a}+1} \]

not realy a challenging one :P:

Answer 14

wow

Answer 15

A straightforward induction also requires using something like this i guess

Answer 16

oo k

Answer 17

Induction step for freckles could be like this :\[\large 2^{2^{a+2}}-1 = \left(2^{2^{a+1}}\right)^2-1 = (2^{2^{a+1}}+1)(2^{2^{a+1}}-1)\]

I don't see a way out if the difference of squares identity doesn't strike...

Answer 18

Fun =D How did you find this, or was the whole thing just supposed to be a tricky way of writing a difference of two squares? xD

Answer 19

Hahaha it just messes with difference of squares ;) saw this neat result in proving \(2^{2^a}+1\) when prime has no primitive roots :

\[2^{2^{a+1}}-1 = \left(2^{2^{a}}\right)^2-1 = (2^{2^{a}}+1)(2^{2^{a}}-1)\]

That means \(2^{2^{a+1}}-1\equiv 0 \pmod{2^{2^{a}}+1}\)
However the number of coprime integers less than \(2^{2^{a}}+1\) would be \(2^{2^a}\) which is always greater than \(2^{a+1}\) proving no primitive roots

Answer 20

:D