Question

Show that if \(a\) and \(b\) are positive integers, then \[\qquad \qquad \large\left(a + \frac12\right)^n +\left(b+ \frac12\right)^n\]is an integer for only finitely many positive integers \(n\).

Answer 1

Still practicing, eh?

Answer 2

Yeah. :/

Answer 3

I have no idea.......(thinks binomial theorem, brain siezes up)

Answer 4

How about this?\[\frac{(2a+1)^n + (2b+1)^n}{2^n}\]

Answer 5

Hmm Binomial, nice thought.

Answer 6

\[\frac{C_0 + C_12a + \ldots C_n(2a)^n + C_0 + C_12b + \ldots +C_n(2b)^n}{2^n}\]

Answer 7

Or, maybe this\[(1+2a)^n + (1+2b)^n = 2^n k, \quad k \in \mathbb{Z}\]

Answer 8

\[2C_0 + 2C_1(a+b) + 4C_2(a^2+b^2) + \ldots +2^nC_n(a^n+b^n)=2^n k \]

Answer 9

I don't think it's gonna help :/

Answer 10

Do you know how to factorize \(x^n + y^n\)?

Answer 11

Hang on a minute, if the result has to be integer how does the 1/2 at the end play out?
(a+b)^n always has b^n at the end

Answer 12

Some other terms have to make it up to an integer.

Answer 13

Not the first term and all the other terms are combinations

Answer 14

And the numerator must be even (per your equation above)

Answer 15

I think I know 2k +1 is always an Odd term. \[\frac{(2a+1)^n + (2b+1)^n}{2^n}\]
Odd^n + Odd^n = Even and Even%2^n = 0, what kind of even numbers have remainder Zero upon division by 2?

Answer 16

For 2^1, every value for a and b should work

Answer 17

For 2^2 or 4, there must be less values.

Answer 18

My head is hurting....:-(

Answer 19

I will try to come back to it later......

Answer 20

Sure, and thanks for trying :-)

Answer 21

One of us under the alias of ManiSarkarCallisto posted this question in M.SE : http://math.stackexchange.com/questions/139035/

Answer 22

Lol, that was me Lololololol

Answer 23

:-)