Can you please help me with the approach for the question?

#1

In the question we are given only 1 side and asked to find the remaining 2 sides which form Pythagoras Triplet but since it is not mentioned that the given input is for the largest side I am not able to derive an approach for the problem.

#2

@anurag.codingsavage If m>n>0 are integers, then (m2-n2, 2mn, m2-n2) is a Pythagorean triple. This is easily seen with a bit of algebra. Thus, plugging in various (m,n) will give various triples.
Now, our goal is to find a triple of the form (m2-n2, 2mn, m2-n2) in which our input value is one of the catheti. Note that the catheti are m2-n2 and 2mn. Also, note that the second cathetus is always even. Thus, it makes sense to consider two cases:

  • If a is even, then let’s try to equate it with the second cathetus, 2mn, i.e., let’s try to find an (m,n) pair such that 2mn = a. In this case, is even, so we can divide by 2. We have mn = a/2. Thus, all we have to do is to find a factorization of a/2 into mn such that m>n. You can try various factoring methods — even naïve O(√a) time method will pass - but in this case , we don’t really even need to do that, since the trivial (m,n) = (a/2,1) works!. This gives us the solution triple ( (a^2)/4 - 1, a, (a^2)/4 + 1 ). Note that (a^2)/4 is an integer since a is even.
  • If a is odd, then we can’t equate it with the second cathetus, so let’s try instead m2-n2, i.e. we want to find an (m,n) such that m2-n2 = a. Node that m2-n2 = (m+n)(m-n), so all boils down again to factoring! Let’s try our trivial factorization again: (m+n, m-n) = (a,1) What we get is a system of two linear equations: m+n=a and m-n=1 which has solution m=(a+1)/2 and n=(a-1)/2. Substituting back, this gives us the triple (a, (a^2 - 1)/2, (a^2 + 1)/2). Note that (a^2 + 1)/2 is an integer as a is odd.
public class pythagoras {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        Scanner scn = new Scanner(System.in);

        long n = scn.nextLong();
        evaluate(n);
    }

static void evaluate(long n) {
        if (n == 1 || n == 2)
            System.out.println(-1);

        else if (n % 2 == 0) {

            // Calculating for even case
            System.out.println(((n * n) / 4) - 1 + " " + (((n * n) / 4) + 1));
        }

        else if (n % 2 != 0) {

            System.out.println((n * n - 1) / 2 + " " + (n * n + 1) / 2);
        }
    }

}