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From: num-curious on 11 Mar 2010 11:19
If p, q, n are integers and pq divides (p+q)^n, it is easy to see either by application of binomial theorem or otherwise that p, q must have a common prime divisor.

Can this result be easily extrapolated when we throw in one more variable into the mix?

i.e, if p, q, r, n are integers and pqr divides
(p+q+r)^n, then is it possible to conclude either by application of multinomial theorem or elementary algebra that p, q, r cannot be relatively prime to each other?


Thanks!
From: Robert Israel on 11 Mar 2010 23:03
num-curious <govegannow(a)gmail.com> writes:

> If p, q, n are integers and pq divides (p+q)^n, it is easy to see either by
> application of binomial theorem or otherwise that p, q must have a common
> prime divisor.

Try p=q=1. But otherwise it is true.

> Can this result be easily extrapolated when we throw in one more variable
> into the mix?
>
> i.e, if p, q, r, n are integers and pqr divides
> (p+q+r)^n, then is it possible to conclude either by application of
> multinomial theorem or elementary algebra that p, q, r cannot be relatively
> prime to each other?

Try p=1, q=1, r=2.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Robert Israel on 12 Mar 2010 00:27
Robert Israel <israel(a)math.MyUniversitysInitials.ca> writes:

> num-curious <govegannow(a)gmail.com> writes:
>
> > If p, q, n are integers and pq divides (p+q)^n, it is easy to see either
> > by
> > application of binomial theorem or otherwise that p, q must have a common
> > prime divisor.
>
> Try p=q=1. But otherwise it is true.
>
> > Can this result be easily extrapolated when we throw in one more
> > variable
> > into the mix?
> >
> > i.e, if p, q, r, n are integers and pqr divides
> > (p+q+r)^n, then is it possible to conclude either by application of
> > multinomial theorem or elementary algebra that p, q, r cannot be
> > relatively
> > prime to each other?
>
> Try p=1, q=1, r=2.

You might also try p=5, q=9, r=16, n=4.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: num-curious on 11 Mar 2010 14:59
> You might also try p=5, q=9, r=16, n=4.

Prof. Israel,

Thanks for your response. How about when n is odd (and > 1)? Do we have similar examples even for that case (i.e, with p, q, r still being relatively prime)?

Thanks again.
From: num-curious on 11 Mar 2010 15:08
Ok, just figured out I could try the same example (p=5, q=9, r=16) with n = 5 or n = 7.

Thanks!
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