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F. to P. generator (Posted on 2014-06-25) Difficulty: 3 of 5
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).

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Isn't this more general and simple? | Comment 2 of 4 |
(In reply to Solution by broll)

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

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

Isn't it now enough to show a+b=c ?
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

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