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 n1 and n+1
The sum of reciprocals:
1/(n1)+1/(n+1)=2n/(n^21)
To show 2n and n^21 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^21)^2 = n^4 + 2n^2 + 1 = (n^2+1)^2

Posted by Jer
on 20150728 10:34:40 