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| Math Central | Quandaries & Queries |
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Given:- {[(n.p)^p] / [c + (n.p)]} = "an integer" (where c > (n.p); c and n are integers, n is even and p is an odd prime number) Can one prove that c cannot be relatively prime to (n.p)? { i.e. c must have the factor (n.p) ?} If so, is the converse also true? {i.e. If c is relatively prime to (n.p), then [c + (n.p)] cannot be a factor of (n.p)^p ?} Lastly but similarly, if c is relatively prime to a and c > a > (n.p) ; (c - a) > (n.p) **; and c is not relatively prime to (n.p) would it be true that (c - a) cannot be a factor (n.p)^p ? I hope my question makes sense? |
Andrew,
if I understand you properly n.p is n 
p and you are looking at (np)p/(c+np) and wants to know about the relation of c to np whenever this ratio is an integer. If this ratio is an integer, every prime q that divides the denominator must divide the numerator. Thus q is p or a divisor of n. If q is in fact p then we must have that p divides c + np which means that p divides c so that c is not relatively prime to np. If q is not p then it divides n and also divides the denominator c + np, but this means that q divides c so that c is not relatively prime to np.
One way to look at the converse (or the above if you like) is to write it as (np)p = k(c+np) for some integer k. Now any divisor of the right hand side must divide the left hand side but if c is relatively prime to np then c+np has a prime divisor q that doesn't divide n or p and thus q cannot divide the left hand side, a contradiction.
Penny
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