Show that:

and

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- December 20th 2008, 10:39 AMSeaDivisibility 13
Show that:

and - December 20th 2008, 11:14 AMMoo
- December 23rd 2008, 03:24 PMJaneBennet
Are you trying to prove the converse, namely

- December 23rd 2008, 04:08 PMSea

I don't know...:(...

But good question...:)...

(Why?) ...write...

http://www.mathhelpforum.com/math-help/number-theory/65693-divisibility-14-a.html

I think...Maybe use binomial theorem... - December 23rd 2008, 04:20 PMJaneBennet
I would suggest maybe using prime factorization.

- December 23rd 2008, 06:23 PMSeaQuote:

I would suggest maybe using prime factorization.

I think;

and

and

I think...? - January 16th 2009, 04:53 AMInvisibleMan
If your'e trying to prove that a^k|b^k => a|b then

a^k=nb^k

(a/b)^k=n

a/b=n^1/k

Now a|b iff n=m^k for some integer m, the problem is to show that it is so.

You can assume by contradiction that a!|b which means there exists p and r such that:

a=pb+r r<=b

Now write a^k=(pb+r)^k and see that it contradicts our assumption.