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  1. Prepare
  2. Mathematics
  3. Algebra
  4. Pythagorean Triple
  5. Discussions

Pythagorean Triple

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31 Discussions

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  • fionawynn
     + 0 comments
    def pythagoreanTriple(a):
    if a % 2 != 0:
        b = (a*a - 1) // 2
        c = b + 1
    else:
        m = a // 2
        b = m*m - 1
        c = m*m + 1
    return a, b, c

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     + 0 comments

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  • masrour_abdelkb1
     + 0 comments

    def pythagoreanTriple(a): if a%2 ==1 : n= (a-1)//2 m=n+1 else : if a % 4 == 0: m = a//4 + 1 n = a//4 -1 else : n=1 m=a//2 return [m**2-n**2,2*m*n,m**2+n**2]

  • shravanatirtha
     + 0 comments

    Complete Java Solution

    import java.io.*;
import java.util.*;
import java.math.BigInteger;


public class PythagoreanTriple
{
	public static void main(String[] args) 
	{
		Scanner sc=new Scanner(System.in);	
		BigInteger b=sc.nextBigInteger();

if(!b.mod(new BigInteger("2")).equals(BigInteger.ZERO))
		{
			//if odd
			BigInteger c=b.pow(2);
			c=c.subtract(new BigInteger("1"));
			c=c.divide(new BigInteger("2"));
			BigInteger a=c.add(new BigInteger("1"));
			System.out.println( b +" "+a +" "+c);
		}
		else
		{
			///if even
		BigInteger c=b.divide(new BigInteger("2"));
			c=c.pow(2); 
			c=c.subtract(new BigInteger("1"));
			
		BigInteger a=c.add(new BigInteger("2"));
			System.out.println( b +" "+a +" "+c);
		}
	}
}

  • sujandas998
     + 0 comments
    1. If a is odd then a^2 is also odd. Now there exists a k, such that, a=2k+1 Therefore a^2=(2k+1)^2=4k^2+4k+1=2b+1, where b=2k^2+2k Hence, a^2=2b+1=(b+1)^2-b^2. i,e, a^2+b^2=c^2, where c=b+1
    2. If a is even then there exists an m such that 2m=a. We know that, (m^2-1)^2+(2m)^2=(m^2+1)^2 , hence a=2m , b=m^2-1 and c=m^2+1.