This problem is pretty: Suppose are in and for all Prove that

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- Jun 25th 2009, 12:14 PMNonCommAlgReals & Integers
This problem is pretty: Suppose are in and for all Prove that

- Jun 26th 2009, 02:16 AMAMI
(1)

(2)

(1),(2) (3)

(1),(3) ; let with

and (4)

(5)

.

Case 1: . This implies

If are odd.

. We have false because it is odd.

We are then left with which means that .

Case 2: . By induction, we prove that :

Fix and suppose but .

(6)

. If false

- contradiction with .

So .

Kind of long.. I hope it's correct though..(Worried) - Jun 26th 2009, 04:08 AMNonCommAlg
- Jun 26th 2009, 05:52 AMAMI
Oh-oh, acted sort of like a prime (sometimes). Sorry. (Blush)

I'll try again, and if it's still not good, then I'll better stop and leave it for someone else.. I'm sorry for wasting your time if I made some mistakes again (Sadsmile)

(1)

(2)

(1),(2) (3)

(1),(3) ; let with

and (4)

I left this part unchanged. And now:

Suppose is even. This implies and odd, hence is odd for every odd . But , so we get for every odd - impossible. This means that must be odd.

. Having and , we get , so .

By induction, we show that .

Suppose . Then, .

.

, where .

.

is fixed arbitrarily and , so .

. - Jun 26th 2009, 10:28 AMSampras
divides for all . Maybe we can use that?

- Jun 26th 2009, 03:37 PMNonCommAlg
the line marked is not quite right because you're fixing an which might have a common divisor with my solution: as you showed suppose and choose a prime divisor of

let be any positive integer such that since we have thus but we know

thus for all positive integers not divisible by which is impossible. - Jun 27th 2009, 12:49 AMAMI
Yes, thank you. And for your clear solution, too.

I guess, since is fixed, I could have chosen to show that only for those with ...

Thanks again, sorry for my mistakes. I'll try to do better next time... (Speechless)