All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Harmonic Integers (Posted on 2006-02-07) Difficulty: 3 of 5
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: 2.2000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution I think I have it | Comment 1 of 5

I started by setting 1/b + k = 1/c, and 1/a +2k = 1/c, then rewrote these in terms of a, and b to get:

a = c/(1-2ck)


then substituted these into the equation a2 + b2 = c2.

After I reorganized the equation I ended up with:

4(ck)4 - 6(ck)3 + 8(ck)2 - 1 = 0. 

Solving for (ck) gave me four solutions.  Two were complex, and the other two were real, but irrational.  Since both c and k must be integers to fit the requirements of the problem, and since the product of two integers can not be irrational nor complex, then the answer to the question posed by K Sengupta must be:

No.   a2 + b2 can not be equal to c2.

Edited on February 8, 2006, 9:19 am
  Posted by Mindrod on 2006-02-07 22:52:15

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information