Fibonacci Day: More Patterns in the Squares

Today yet again is a Fibonacci Day! It is October 8th, and 8 is the sixth Fibonacci number. To keep the theme of last week, let's use the square Fibonacci numbers again. Here they are:

1     1     2     3     5     8     13     21     34
1     1     4     9   25   64   169   441 1156

Let's take each Fibonacci number and move one away from it. Now, we'll multiply those numbers and see how close we get to the square.

1) 0 x 1 = 0 = 1^2 - 1
2) 1 x 2 = 2 = 1^2 + 1
3) 1 x 3 = 3 = 2^2 - 1
4) 2 x 5 = 10 = 3^2 + 1
5) 3 x 8 = 24 = 5^2 - 1
6) 5 x 13 = 65 = 8^2 + 1
7) 8 x 21 = 168 = 13^2 - 1

And so on and so forth. Basically, Fn-1 x Fn+1 = Fn^2 ± 1, or even more accurate, Fn-1 x Fn+1 = Fn^2 + (-1)^n. Let's try it again, this time looking two away from the number. Keep in mind that 1 is the negative-first Fibonacci number.

1) 1 x 2 = 2 = 1^2 + 1
2) 0 x 5 = 0 = 1^2 - 1
3) 1 x 5 = 5 = 2^2 + 1
4) 1 x 8 = 8 = 3^2 - 1
5) 2 x 13 = 26 = 5^2 + 1
6) 3 x 21 = 63 = 8^2 - 1
7) 5 x 34 = 170 = 13^2 + 1

This time, we have pretty much the same pattern. Fn-2 x Fn+2 = Fn^2 ± 1, or  Fn-2 x Fn+2 = Fn^2 - (-1)^n. How about we move three away. It's the same type of pattern, but a little different. Keep in mind that -1 is the negative-second Fibonacci number (since a Fibonacci number is the two numbers before it added together, than the zero comes from x + 1, or -1 + 1).

1) -1 x 3 = -3 = 1^2 - 4
2) 1 x 5 = 5 = 1^2 + 4
3) 0 x 8 = 0 = 2^2 - 4
4) 1 x 13 = 13 = 3^2 + 4
5) 1 x 21 = 21 = 5^2 - 4
6) 2 x 34 = 68 = 8^2 + 4
7) 3 x 55 = 165 = 13^2 - 4

We have the same idea. We are stuck with a four, giving us the pattern of  Fn-3 x Fn+3 = Fn^2 + 4(-1)^n. Let's look at our neighbors four away and see if we can see the pattern better. What do you think the negative-third Fibonacci number is? If you got two, then good job.

1) 2 x 5 = 10 = 1^2 + 9
2) -1 x 8 = -8 = 1^2 - 9
3) 1 x 13 = 13 = 2^2 + 9
4) 0 x 21 = 0 = 3^2 - 9
5) 1 x 34 = 34 = 5^2 + 9
6) 1 x 55 = 55 = 8^2 - 9
7) 2 x 89 = 178 = 13^2 + 9

Same idea again. We have Fn-4 x Fn+4 = Fn^2 - 9(-1)^n. However, the four and nine aren't there randomly. Let's look these differences closer.

1, 1, 4, 9

Recognize them? They are the squares of the Fibonacci numbers again! If you go five away, it is the square of the fifth Fibonacci number, six away is the square of the sixth Fibonacci number, one hundred away is the square of the hundredth Fibonacci number. Basically, a general formula is Fn-a x Fn+a = Fn^2 ± (Fa^2)(-1)^n. Or, you can use the below formula to be even more accurate:

Fn-a x Fn+a = Fn^2 - ((-1)^a)(Fa^2)((-1)^n)

I have no clue why this works, but please put up a proof if you know it. This is one of the coolest things about Fibonacci numbers!