Fibonacci Proof: No Consecutive Numbers Divisible by Integers | Helpful Hints

[chocok]

I got stuck on proving no two consecutive Fibonacci numbers are divisible by any integer greater than 1.

Some hint please?

 

[matt grime]

If d divides A and B it divides A-B, that is sufficient for you to be able to prove this result.

 

[tacman]

Sure, here are some helpful hints to guide you in your proof:

1. Start by assuming that there exists two consecutive Fibonacci numbers, F(n) and F(n+1), that are both divisible by some integer m > 1. This means that F(n) = km and F(n+1) = lm, where k and l are integers.

2. Use the definition of Fibonacci numbers, which states that F(n) = F(n-1) + F(n-2). Substituting in our assumed values for F(n) and F(n+1), we get km = F(n-1) + F(n-2) and lm = F(n-1) + F(n-2).

3. Subtract the two equations to eliminate F(n-1), leaving us with (k-l)m = F(n-2). This means that F(n-2) is also divisible by m.

4. Continue this process by substituting in the definition of Fibonacci numbers again, F(n-2) = F(n-3) + F(n-4), and repeating the same steps. Eventually, you will reach a point where F(2) and F(1) are both divisible by m, since the Fibonacci sequence starts with 1 and 1.

5. However, this is a contradiction because F(2) = 1 and F(1) = 1, which means that m must equal 1. But we initially assumed that m was greater than 1, so our assumption was incorrect.

6. Therefore, our initial assumption that there exists two consecutive Fibonacci numbers divisible by m > 1 is false, and we have proven that no two consecutive Fibonacci numbers are divisible by any integer greater than 1.

I hope these hints help you in your proof! Remember to clearly state your assumptions, use the definition of Fibonacci numbers, and follow the logic until you reach a contradiction. Good luck!