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 Isn't this more general and simple? by Jer)

Hah! I started with just that idea in mind, but then got distracted. You are correct; thanks for putting me back on the right track.

The generalisation I was originally aiming for also involves Fibonacci numbers F_(n), where a and b can be anything:

2*(F_(n-1)a+F_(n)b)*(F_(n)a+F_(n+1)b)       = x
(F_(n-2)a+F_(n-1)b)*(F_(n+1)a+F_(n+2)b)   = y
(F_(n-1)a+F_(n)b)^2+(F_(n)a+F_(n+1)b)^2  = z

then x^2+y^2=z^2.

Small values of c:

a^2+2ab+2b^2, 2a^2+6ab+5b^2, 5a^2+16ab+13b^2, 13a^2+42ab+34b^2, 34a^2+110ab+89b^2,..etc; generally:

F_(n-1)a^2+2F_(n)ab+F_(n+1)b^2 .

When a=b=1, these expressions return the Fibonacci numbers F_(2n-1) {5,13,34,89,...} themselves.

Very pretty.

Edited on June 27, 2014, 4:03 am

Posted by broll on 2014-06-27 02:09:59