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?
See The Solution | Submitted by K Sengupta |
Rating: 4.5000 (2 votes) |
re: One Approach |
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(In reply to One Approach by Richard)
Once we have 2 a^2 = b^2, there are, as Richard correctly observes, various ways forward to prove this has no solution in positive integers.
The simplest I think is to note that it implies
the irrational sqrt(2) = b/a rational.
Posted by goFish on 2006-02-21 16:04:01 |