Proof (Fibonacci numbers, irrational)?

Question

Prove/extend or disprove/salvage

(x) denotes subscript
Statement:
F(0)= 1 F(1)= 1 F(2) = 2 F(3) =3 .... denotes the fibonaccia numbers. Then for n > 1, (F(n)F(n+2))^1/2 is irrational.

And does a similar result hold for the number (F(n))^1/2

Answer 1

For the first part we use the identity

F(n)F(n+2) - F(n+1)^2 = +/- 1 so that F(n)F(n+2) is
within 1 of the value F(n+1)^2 and therefore is not the square
of a whole number since squares do not differ by 1.
Therefore {(F(n)F(n+2)}^(1/2) is irrational.

You can get Cohn's result if you are manic enough and
want to use the identities,

F(2n+1) = F(n+1)^2 + F(n)^2,

F(2n) = F(n+1)^2 + F(n-1)^2,

so that if F(N) is a square, then it is also a pythagorean
triple and relatively prime to the squared fibonacci numbers whose sum is F(N). Then , if you assign the pythagorean
triplet values you get F(N) = u^2 + v^2, and if N is odd, for example, and you assign f(n+1) = 2uv,
then f(n+1) and f(n) are assigned the values 2uv, u^2-v^2
from which you can get f(n-1) = f(n+1) - f(n) = 2uv-(u^2-v^2),
Eventually you get f(1) = a fourth degree poly G( u,v) = 1
or you may go to zero which is f(2)-f(1) = 0 = G(u,v). The
point is that you get managable diophantine equations in
u and v which have no integer solutions except for the possibilities which allow for the squares, F(1),F(2), and F(12)

The entire proof is less than a page.

Answer 2

I'll do the second part first.
F(n)^½ is rational when and only when
n = 1, 2 or 12,
with F(n) = 1,1, and 144.
These are the only squares in the Fibonacci
sequence. This was proved by J.H.E. Cohn
in 1964.
Cohn's result also answers question 1 as follows.
One can prove that
(F_m,F_n) = F_(m,n)
where (a,b) denotes the GCF of a and b.
So (F_n,F_(n+2) = F_(n,n+2) = F_1 or F_2 = 1
So F_n*F_(n+2) = y²
implies both F_n and F_(n+2)
are squares.
But F_n is a square only for n = 1,2 and 144
and in none of these cases is F_(n+2) a square.
So your first proposition is true, even if n = 1 because
F_1*F_3 = 2.

Answer 3

APEX; ~The fourth Fibonacci selection is 3. each and every fourth Fibonacci selection after it is gently divisible by skill of three. ~beginning with F11, the version between any Fibonacci selection and the Fibonacci selection that comes 10 in the past it (as an occasion, F37 - F27) is a selection gently divisible by skill of 11. ~The sum of the 1st n Fibonacci numbers is often precisely a million decrease than yet another Fibonacci selection.