Added to JS, Python, and Java.

The formula is not accurate for n > 71. Hence, this solution won't work for even trivial values.

can you tell me what do you mean by the #2 and #3 flaws? I wanna know the bottlenecks they could cause us?

Fun! But it's not quite right! For example your version of productFib(24) gives (4,6,true), when 4 and 6 aren't fibonacci numbers. Other examples are 60, 77, 135, 160, and 198. These are the product of two numbers that very closely have a golden ratio between them, but they don't appear in the fibonacci sequence.

See https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression

phi=(sqrt(5)+1)/2 = 2/(sqrt(5)-1) ~= F(n+1)/F(n) for large n. So F(n) ~= F(n+1)/phi, where by ~= I mean "approximately equals". It turns out the round function can be used since F(n+1)/phi will be within 0.5 of F(n).

I never thought to use an array in a for loop before, what an intiguing use.

i've added == in the description's JS code blocks

why sqrt(5)?

Good point, calling the object "arr" confused me.

It is definetely not a byproduct of the kata requiring the next highest pair. The reason people haven't been using this solution is because most don't know about it.

I know this is old, but thanks for sharing! Very interesting.

I'm forced to reply to the wrong person, but here we go

Brute force generally means to generate every possible combination until you find one that gives you an answer. In this case, it refers to starting from the bottom of the fibonacci sequence and progressing through it until you arrive at a stopping case (fib(n) * fib(n+1) >= prod).

Hey! I'm new here. Can you please explain what you mean by brute force algorithm?

Thank you for your suggestion!
I saw your solution, but it was not actually working as I expected, so I decided to solve it like this:
https://www.codewars.com/kata/reviews/554398d646002df491000183/groups/5dc2fa1a83a99000015a24ba