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From: num-curious on 11 Mar 2010 15:52
Changing the conditions a bit:

1. p, q, r are distinct integers (assume all > 0).
2. n is odd > 1
3. (p) > (q+r)
4. (npqr)(2^n) divides (p-q-r)^n

Is (4) possible when p, q, r are relatively prime? If not, can it be proved that it is not possible using multinomial theorem or any elementary algebra?

Thanks!
From: Rob Johnson on 12 Mar 2010 02:28
In article <1204048055.366362.1268360386588.JavaMail.root(a)gallium.mathforum.org>,
num-curious <govegannow(a)gmail.com> wrote:
>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?

Assume that pq divides (p+q)^n for some n.

Suppose that d is a prime that divides pq. By our assumption, it
follows that d divides (p+q)^n. Since d is a prime, we must have
that d divides (p+q). Since d divides pq, either d divides p or
d divides q. If d divides p, then d divides (p+q)-p = q. If d
divides q, then d divides (p+q)-q = p. Thus, if d divides pq, then
d divides both p and q.

Therefore, any prime that divides pq must also divide (p,q). The
only way that (p,q) = 1, is if pq = 1.

For more than 2 variables, try p = 2, q = 3, r = 25, and n = 2.

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
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From: Robert Israel on 12 Mar 2010 02:35
num-curious <govegannow(a)gmail.com> writes:

> Changing the conditions a bit:
>
> 1. p, q, r are distinct integers (assume all > 0).
> 2. n is odd > 1
> 3. (p) > (q+r)
> 4. (npqr)(2^n) divides (p-q-r)^n
>
> Is (4) possible when p, q, r are relatively prime? If not, can it be proved
> that it is not possible using multinomial theorem or any elementary
> algebra?

p=27, q=1, r=2, n=9

p=16, q=1, r=3, n=9

p=25, q=1, r=4, n=5

....
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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