Take any 4 consecutive Fibonacci numbers, say d,f,g,h.
Evaluate:
a=2*f*g;
b=d*h;
c=f²+g².

Prove that a triangle with sides a,b,c is a right-angle triangle (for any quadruplet of successive Fibo numbers).

(In reply to Solution by broll)

Let f=u,d=v, g=u+v, h=2u+v

a=2*f*g = 2u(u+v) = 2u² + 2uv
b=d*h = v(2u+v) = 2uv + v²
c=f²+g² = u²+ (u+v)² = 2u² + 2uv + v²

Isn't it now enough to show a²+b²=c² ?
It's simple enough to show that I won't type out the algebra.  The point it this rule works for any quadruplet numbers where the sum of the first two is the third and the sum of the second and third is the fourth.  The Fibonacci sequence is just a specific sequence.

Please correct me if I'm wrong.

Posted by Jer on 2014-06-26 11:07:00