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November 10th, 2013, 03:06 AM   #1
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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.
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November 10th, 2013, 03:14 AM   #2
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Re: Pythagorean triples and unit fractions

I think it is square of n^2 + 2n +2
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November 10th, 2013, 03:57 AM   #3
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Re: Pythagorean triples and unit fractions

That's correct. Thanks. I would have liked the intermediate steps to the factorization.
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November 10th, 2013, 04:11 AM   #4
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It's obvious the equal factors would be quadratic. What would their constant term be?
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