Find Pythagorean triplet for which a + b + c = 1000

Question

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a² + b² = c²

For example, 3² + 4

Answer 1

I know this question is quite old, and everyone has been posting solutions with 3 for loops, which is not needed. I got this solved in O(n), by **equating the formulas**; **a+b+c=1000 and a^2 + b^2 = c^2**

So, solving further we get;

a+b = 1000-c

(a+b)^2 = (1000-c)^2

If we solve further we deduce it to;

a=((50000-(1000*b))/(1000-b)). We loop for "b", and find "a".

Once we have "a" and "b", we get "c".

public long pythagorasTriplet(){
    long a = 0, b=0 , c=0;

    for(long divisor=1; divisor<1000; divisor++){
        if( ((500000-(1000*divisor))%(1000-divisor)) ==0){
            a = (500000 - (1000*divisor))/(1000-divisor);
            b = divisor;
            c = (long)Math.sqrt(a*a + b*b);
            System.out.println("a is " + a + " b is: " + b + " c is : " + c);
            break;
        }
    }
    return a*b*c;
}