We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Mathematics
  3. Algebra
  4. Pythagorean Triple
  5. Discussions

Pythagorean Triple

  • 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.