2015/068) find parametric form of Pythagorean triplet for $(3,4,5), (5,12,13) \cdots$ that is difference between hypotenuse and one leg is 1

that means we need to find $(x,y,y+1)$ where

$x^2 + y^2 = y^2 + 2y + 1$

or $x^2 = 2y + 1$

now x has to be odd say $2n + 1$

so $x^2 = 4n^2 + 4n + 1 = 2y + 1$

or $y = 2n^2 + 2n$

so the sides are $( 2n+ 1, 2n^2+2n, 2n^2+2n+ 1)$

by putting n = 1,2 ,3 we get the sequence