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Highest Common Factors
By Kate Graham on Monday, June 03, 2002 - 08:37 pm:

I can't get my head around a proof for the following problem.
If a,b and c are positive integers and p is prime and hcf{a2p,b}=p2, then hcf{a2,b}=p2.
Any ideas on where to go with it would be appreciated.

By David Loeffler on Tuesday, June 04, 2002 - 10:42 pm:

p2 divides a2p, so p divides a2. Hence p divides a, so p2 divides a2. Also we know p2 divides b. Hence p2 divides a2 and b. And any larger common factor of these two also divides a2p, b, contradicting the hypothesis, so p2 is the highest common factor.

David

By Kate Graham on Wednesday, June 05, 2002 - 03:01 pm:

Thanks David, I just couldn't get started on it and the annoying thing is I knew I knew it. Know what I mean?
Thanks again for saving my sanity!!! LOL