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Right Triangle (Posted on 2003-08-05) Difficulty: 3 of 5

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.

See The Solution Submitted by np_rt    
Rating: 3.4000 (5 votes)

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Hints/Tips First ideas | Comment 1 of 12
By multiplying each side by a²b²c² and subtracting a²c² from each side, then factoring, you can get a²(b²-c²) = b²c²

From this you can see that the difference of two perfect squares has to be a factor of their product.
  Posted by Gamer on 2003-08-05 12:07:08
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