This problem is pretty: Suppose are in and for all Prove that
(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..
Oh-oh, acted sort of like a prime (sometimes). Sorry.
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
(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 .
.
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.