Try to solve this problem by a method which is different from the solution to "Harmonic Integers" problem.
Given that, a, b, and c are all positive integers so that a < b < c, and 1/a, 1/b, and 1/c are in Arithmetic Progression, can a + b be equal to c?
One Approach
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The harmonic condition is 1/b-1/a=1/c-1/b which upon using a common
denominator yields 2*a*c-b*c-a*b=0. The condition c=a+b when
substituted then yields 2*a^2=b^2, which is impossible by unique
factorization.
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Posted by Richard
on 2006-02-21 11:22:39 |
