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- Dec 19th 2008, 02:03 AMSeaDivisibility (gcd) 8
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- Dec 19th 2008, 02:40 AMIsomorphism
Let and

Thus . -----------------(1)

By Bezout's identity we know that there exists non negative integers x and y such that .

But

Thus -------------(2)

(1) and (2) imply - Dec 19th 2008, 03:17 AMSea
I don't know...

(Why?)

(Why?) - Dec 19th 2008, 03:29 AMMoo
- Dec 19th 2008, 04:53 AMPaulRS
A slightly different way.

We have (as Moo points out) thus (1)

Now let be a common divisor of and i.e.

From the first congruence we get and from the second thus it must be that hence

So every common divisor divides and by (1) we are done (because it's the greatest possible)