I figured out a solution for problem Div2C. Given a number n , we have to find its other two Pythagorean Triplets.

My approach was as follows:

I stored the factors of n in a vector. Since we have n*n to deal with, I basically decomposed n*n into all possible combinations of any two of (n*n)'s factors p,q such that p*q = n*n

now, a*a +n*n = c*c

=> n*n = (c-a)*(c+a) = p*q

Now assuming min(p,q) = c-a and max(p,q) = c+a, I just checked if the Pythagorean equation is satisfied. My code

What I still can't figure out in my own code is how my solution gives a correct answer if n is a hypotenuse. I have not handled the case x*x + y*y =n*n. My code seems to prove that if x*x + y*y = n*n, then there always exists a,b such that a*a + n*n = b*b. Simply speaking, hypotenuse of any right angled triangle is always cathetus of some other right triangle. Does this always hold true or is there a counter case. A proof from someone would be really helpful. Thanks :)