On Sun, 23 Dec 2007 15:15:04 -0500, "Gary S. Simon"
<garscosi@[EMAIL PROTECTED]
> wrote in
<news:garscosi-35F51E.15150423122007@[EMAIL PROTECTED]
>
in alt.algebra.help:
> Riding the subway home on the last leg of our return from
> the Staten Island Zoo yesterday, my seven year old son
> Dan asked me to do the following (Dan doesn't know much
> terminology, so I'm including it parenthetically):
> 1. Pick a Fibonacci number (Fn)
> 2. Square it
> 3. Square the Fibonacci "two before the one you picked" (F(n-2))
> 4. Subtract it from the first square
> 5. You get the Fibonacci when you add the other two (F(2n-2)).
This can be shown fairly easily using the well-known Binet
formula for the n-th Fibonacci number. For brevity let
s = sqrt(5), x = (1 + s)/2, and y = (1 - s)/2; then F(n) =
(x^n - y^n)/s. Note that xy = -1, x + y = 1, and x - y = s.
Now
F(n)^2 - F(n-2)^2 =
[(x^n - y^n)^2 - (x^(n-2) - y^(n-2))^2]/5 =
[x^(2n) - 2*(xy)^n + y^(2n) - x^(2n-4) + 2*(xy)^(n-2) - y^(2n-4)]/5 =
[x^(2n) - 2(-1)^n + y^(2n) - x^(2n-4) + 2(-1)^(n-2) - y^(2n-4)]/5 =
[x^(2n) + y^(2n) - x^(2n-4) - y^(2n-4)]/5,
since (-1)^n = (-1)^(n-2).
F(2n-2) = [x^(2n-2) - y^(2n-2)]/s, so we want to show
that
[x^(2n) + y^(2n) - x^(2n-4) - y^(2n-4)]/5 =
[x^(2n-2) - y^(2n-2)]/s,
i.e., that
x^(2n) + y^(2n) - x^(2n-4) - y^(2n-4) =
s * [x^(2n-2) - y^(2n-2)].
Now y = -1/x, so 1/x^2 = y^2, and hence
x^(2n) - x^(2n-4) =
x^(2n-2)*(x^2 - 1/x^2) =
x^(2n-2)*(x^2 - y^2) =
x^(2n-2)*(x + y)*(x - y) =
x^(2n-2) * s.
Similarly,
y^(2n) - y^(2n-4) =
y^(2n-2)*(y^2 - 1/y^2) =
y^(2n-2)*(y^2 - x^2) =
y^(2n-2)*(y + x)*(y - x) =
y^(2n-2) * (-s).
Thus,
x^(2n) + y^(2n) - x^(2n-4) - y^(2n-4) =
x^(2n) - x^(2n-4) + y^(2n) - y^(2n-4) =
x^(2n-2) * s + y^(2n-2) * (-s) =
s * [x^(2n-2) - y^(2n-2)],
as desired.
£Brian
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