Why it only passes only 1 test case what changes should i do in my code

#1

#include<bits/stdc++.h>

using namespace std;

int main() {
int N;
cin>>N;
if(N<2){
cout<<-1;
}
if(N%2==0){
int m=N/2;
int n=1;
cout<<mm-nn<<" “<<mm+nn<<endl;
}
else{
int m=N+1/2;
int n=N-1/2;
cout<<2mn<<” "<<mm+nn;
}
return 0;
}

#2

@rathidevesh906_1dcc73a061c159bb, refer here https://ide.codingblocks.com/s/657506

reasoning :-
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.
#3

@rathidevesh906_1dcc73a061c159bb, u can also refer https://www.geeksforgeeks.org/generate-pythagoras-triplet-single-integer/

#4

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