The sum of the squares of consecutive Fibonacci numbers is another Fibonacci number. Specifically we have
and so we have the following right triangle.
The hypotenuse will always be irrational because the only Fibonacci numbers that are squares are 1 and 144, and 144 is the 12th Fibonacci number.
It looks like the only Pythagorean triple of this sort is (0,1;1).
https://math.stackexchange.com/questions/1012999/square-fibonacci-numbers
What does F2n+1 the hypotenuese notation mean?
Nah it’s observed by all combinations.
0^2 + 1^2 = 1
1^2 + 1^2 = 2
1^2 + 2^2 = 5
2^2 + 3^2 = 13
…
You always get the odd-numbered numbers in the Fibonacci sequence.
The subscripts are the index of the Fibonacci numbers, so F_n is the nth Fibonacci number.
Interesting and bizarre.
Is there an explanation or proof of this phenomenon somewhere?
Here is a long but explicit proof.
Here’s a sketch of a more elegant proof via Dmitry Rubanovich: “F_{n+m}=F_n F_{m-1} + F_{n+1} F_m. Now set m=n+1 and you get the proof.” See the discussion here.
Interesting
The square of any number in a Fibonacci sequence is equal to the square of its previous number plus the square of its posterior divided by its before-anterior, this is the same as Pythagoras’s theorem for the right triangle! It’s the beauty of algebra finding geometry!
The beginning of the Fibonacci sequence (0,1,1,2) then shows us that the existence of rational numbers is linked to the need for the ethereal 0 (ZERO) to exist. Since 2/0 does not exist, the equation: (1 ^ 2) = ((1 ^ 2) + (2/0) ^ 2) can only be solved by discovering the square root of 2, the first irrational number since there could be no right triangle without hypotenuse.