For any primitive Pythagorean triplet the following rules apply :
1. The sum of the hypotenuse and one of the legs of a PPT is a square of an odd number,
2. The arithmetic mean of the hypotenuse and the odd leg is a perfect square.

Prove it, please.

Let the 3 sides be A = u^2-v^2, B = 2uv, C = u^2+v^2 (Euclid)
Then B+C=u^2+2uv+v^2=(u+v)^2. Since the triplet is primitive, both A and C must be odd; hence u and v are of opposite parities and  (u+v)^2 is also odd. (1)
And (A+C)/2=(u^2-v^2+u^2+v^2)/2=(2u^2)/2, namely u^2.(2)

Edited on October 6, 2012, 11:47 pm

Posted by broll on 2012-10-06 14:53:18