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November 10th, 2013, 03:06 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  Pythagorean triples and unit fractions
1/3 + 1/5 = 8/15 and 8^2 + 15^2 = 17^2 In general (it appears), the sum of two unit fractions (numerator = 1) whose denominators differ by two, yields a fraction whose numerator and denominator are part of a Pythagorean triple (numerator^2 + denominator^2 = c^2). Prove it! I tried 1/n + 1/(n+2) and got numerator n + (n + 2) and denominator n(n + 2). Adding their squares, we get (n^4) + 4(n^3) + 8(n^2) + 8(n) + 4. I tried showing the expression above is a perfect square, but I couldn't. Thanks. 
November 10th, 2013, 03:14 AM  #2 
Member Joined: Nov 2012 Posts: 61 Thanks: 0  Re: Pythagorean triples and unit fractions
I think it is square of n^2 + 2n +2 
November 10th, 2013, 03:57 AM  #3 
Senior Member Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18  Re: Pythagorean triples and unit fractions
That's correct. Thanks. I would have liked the intermediate steps to the factorization.

November 10th, 2013, 04:11 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,105 Thanks: 2324 
It's obvious the equal factors would be quadratic. What would their constant term be?


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