Pythagoras triplet

#1

why the code is not showing any output although it compiled succesfully.

#2

hello @lovely

check ur loop conditions they are not correct.

also u dont have to use loop for answer , u can get it in o(1) per input.

here is the approach->
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.

code->

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
#define ll long long int
int main(){
    ll x;

    while( cin >> x){
        if(x < 3){ printf("-1\n");  continue; }

        if(x & 1)  printf("%lld %lld", (x*x-1)/2, (x*x+1)/2);
        else{

            printf("%lld %lld", ((x*x)/4)-1, ((x*x)/4)+1);
        }
    }
    return 0;
}
#3

i could not understand this code.can u please tell some other method to solve

#4

check now , this should make sense->

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
#define ll long long int
int main(){
    ll x;

    while( cin >> x){  // we r using this loop becuse we need to process unknown number of input

        if(x < 3){  cout<<"-1\n";  continue; }

        if(x %2 !=0)   cout<<(x*x-1)/2<<" "<<(x*x+1)/2;
        else{

             cout<<((x*x)/4)-1<<" "<< ((x*x)/4)+1;
        }
    }
    return 0;
}
1 Like
#5

while( cin >> x) can we use while(1) instead of this and then inside the while loop cin>>x and updating x by x++?

#6

no

we cannot, why x++ ? x is data only doing x++ will not help.

while(cin>>x) this will stop when no input is left . there is no other way to guess where to stop

#7

but how will we guess it that we have to use this :thinking:

#8

whenever u are required to read unknown number of integers u can use it .

1 Like
#9

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