Take two integers that differ by two, e.g. as 3 and 5.
Create their reciprocals and add them: 1/3 + 1/5 = 8/15.
The numerator and the denominator in the resulting sum are always two numbers of a Pythagorean triplet!
In our case the Pythagorean triplet is 8, 15, 17.

Explain why it always works.

As a problem this is just basic algebra, but the trick is really neat.

Call the integers n-1 and n+1

The sum of reciprocals:
1/(n-1)+1/(n+1)=2n/(n^2-1)

To show 2n and n^2-1 are two numbers of a Pythagorean triplet (specifically the legs) we sum their squares and show the result is a perfect square:
(2n)^2 + (n^2-1)^2 = n^4 + 2n^2 + 1 = (n^2+1)^2

Posted by Jer on 2015-07-28 10:34:40