For readers interested in pythagorean triplets with Consecutive natural numbers, take any odd number -- say 11 -- and square It. 11 squared = 121 = 2(60)+ 1. Add 60 squared to both sides to get 11 squared + 60 squared = 60 squared + 2(60) + 1 = (60 + 1) squared = 61 squared. What do we get? (11, 60, 61). Works every time. Prove it For a general case.
RE: For readers interested in pythagorean triplets with Consecutive natural numbers, take any odd number...
Two numbers, say m & n, form Pythogeran triplets, if n=11 and m=60 First condition n^2 + m^2 = (m+1)^2 11^2+60^2=(60+1)^2 Second condition n^2=2m+1 11^2=2(60)+1 So if we prove first condition to second the this must be true. Consider Second condition n^2=2m+1 add m^2 on both sides n^2+m^2=m^2+2m+1 n^2+m^2=(m+1)^2
