Why is the difference of consecutive primes from Fibonacci sequence divisible by $4$?

Question

The primes represented in the Fibonacci sequence are written in the form $6n + 1$ and $6n -1$, respectively.

$$5=6\times1-1$$ $$13=6\times2+1$$ $$89=6\times15-1$$ $$233=6\times39-1$$ $$1597=6\times266+1$$ $$28657=6\times4776+1$$ $$514229=6\times85705-1$$ $$433494437=6\times 72249073-1$$ $$2971215073=6\times495202512+1$$

This is the part I don't know how to prove.
I have found that $n$ is an odd number when the prime number is written in the form $6n -1$ and $n$ is an even number when the prime number can be written in the form $6n + 1$.

If the above is true than it is clear that diffrence between two prime numbers from Fibonacci sequence bigger than $4$ is divisible with $4$.

I wrote down all possible forms of the differences of two primes.

n and m are even

$6n + 1 - (6m + 1) = 6n + 1 - 6m - 1 = 6n - 6m = 6 (n - m) = 2 \times 3 \times (n - m)$ Since n and m are even numbers, their difference is also even, which means that the difference of the two prime numbers in this case is divisible by 4.

n even, m odd

$6n + 1 - (6m - 1) = 6n + 1 - 6m + 1 = 6 (n + m) +2 = 2 (3 (n + m) + 1)$

Since $n$ is an even number and $m$ is an odd number, their sum is also an odd number. Three times an odd number is an odd number, and adding $1$ to it gives an even number. This proves that the difference is again divisible by $4$.

n odd, m even

$6n - 1 - (6m + 1) = 6n - 1 - 6m - 1 = 6 (n - m) - 2 = 2 ( 3(n - m) - 1) $

If $n$ is an odd number and m is an even number, their difference is an odd number. As in the previous example, three times an odd number is an odd number and adding one to it gives an even number. So the difference is divisible by $4$.

n odd, m odd 

$6n - 1 - (6m - 1) = 6n - 1 - 6m + 1 = 6 (n - m)$
If both numbers are odd, their difference is an even number, which again means that the difference is divisible by $4$.

Answer 1

To expand on the comments:

All Fibonacci primes $>3$ are of the form $4k+1$.

To prove this, suppose it were otherwise. That is, suppose we had an index $n$ for which $F_n\equiv 3 \pmod 4$. We wish to show that $F_n$ is composite.

But, The sequence $\{F_n\}$ is periodic $\pmod 4$ with cycle of length $6$. $\{0,1,1,2, 3, 1\}$. Thus, $F_n\equiv 3\pmod 4\implies n\equiv 4\pmod 6$. In particular, $n$ must be even. Hence, $n$ is not a prime, which implies that $F_n$ is not a prime (other that $n=4$) since $n=2k\implies F_k\,|\,F_n$ and $F_k>1$ if $k>2$.

Note: I expected this to be a duplicate, but was unable to find a match. If somebody can produce a duplicate, I'll delete this.

Answer 2

There are plenty of relations with Fibonnaci numbers, for example $$\forall n\geq 1,F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n(F_{n+1}+F_{n-1})$$which explain @lulu's answer.

From his response, we know that

• $F_{2n}$ is not prime;
• The remainders when you do Euclidean division by 4 are $$0,1,1,2,3,1,\color{red}{0,1,1,2,3,1,},0,1,1,2,3,1,...$$

Let's apply this last idea when you do the Euclidean division by $6$: we obtain a periodicity of $24$

$$0,\boxed{1,1},2,3,\boxed{5},\color{blue}{2,\boxed1,3,4,\boxed{1,5}},0,\boxed{5,5},4,3,\boxed{1},\color{blue}{4,\boxed5,3,2,\boxed{5,1}}$$

Prime Fibonnaci numbers have necessarly remainders $1$ or $5$, as you said it.

There you have it, you have enough information to conclude that what you have said is true :

$\boxed{\text{$n$ is an odd number when the prime number is written in the form $6n -1$}}$

$\boxed{\text{$n$ is an even number when the prime number can be written in the form $6n + 1$}}$