Given a right triangle with lengths that are reciprocals of integers, what is the smallest possible sum of these the integers?

In other words, given a right triangle with lengths 1/a, 1/b, and 1/c, where a, b, and c are all integers, what is the lowest value of a+b+c? Also, prove it.

Taken from CAML, which did not ask for a proof.

There are usually people who have the same question but feel too silly to ask. I don't know what the problem is unless you tell me! :)

I actually thought I wrote why and didn't. Sorry about that...

(I think you mean a² + b² = (ab/c)²) We can conclude that (ab/c)² is an integer because the left side has to be an integer. I was thinking the only thing that could be squared and become an integer is the square root of something (which would be irrational) or another integer. Based on (ab/c) is an integer, (ab/c)² must be a perfect square, and this means the left side (a² + b²) is a perfect square as well.

Posted by Gamer on 2003-08-05 16:57:23