Prove that every other Fibonacci number from 5 on (i.e. 5,13,34,89... etc) can denote the length of a hypotenuse of a right triangle with integer sides.
The way that the Fibonacci numbers perform the trick described above is quite interesting.
1. Let F(n) denote the nth Fibonacci number.
Starting with the well known result: F(n)^22F(n1)^22F(n2)^2 + F(n3)^2 = 0, and rearranging: F(n)^2 + F(n3)^2 = 2F(n1)^2 + 2F(n2)^2.
But as is also well known, (F(n))^2 + (F(n+1))^2 = F(2n+1), and rearranging that slightly, (F(n1))^2 + (F(n2))^2 = F(2n1).
It follows at once that (F(n))^2+(F(n+3))^2=2*F(2n1).
Let (F(n))^2=a, let (F(n+3))^2=b and let F(2n1) = c. Algebraically, a^2+b^2=2c. [1]
So what is c^2?
The first few Fibonacci numbers and their indices are as follows
n F(n)
0 0
1 1
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55
Taking 5 for example: 5 is F(5).
5=1+4, the sum of the squares of 1 and 2, namely F(2) and F(3), but also, as already mentioned, half the sum of 1^2+3^2, namely F(1) and F(4). The square of 5 is 25, and 25 = 3^2+4^2. The smaller of these numbers, 3 = 3*1, or F(4)*F(1). The larger is 4, which is 51^2, or F(5)(F(1))^2
610 is a larger example: 610 is F(15).
610 = 169+441, the squares of F(7) and F(8), and also half the sum of the squares of F(6) and F(9).The square of 610 is 372100, the sum of the squares of 272 and 546. 272 = 34*8, or F(9)*F(6). 546 = 6108^2 or F(15)(F(6))^2.
And so on.
Generally, F(2n1)^2 = (F(n)*F(n+3))^2+(F(2n1)(F(n))^2)^2.
Algebraically again, c^2=(ab)^2+(cb^2)^2, a particularly neat expression, because it simplifies to 2c=a^2+b^2 [1], as was to be shown.
Edited on August 7, 2018, 3:00 am

Posted by broll
on 20180807 02:39:31 