For $n\ge4,$ prove that $F_n+1$ is not prime, where $F_n$ is $n^{th}$ Fibonacci number.

Question

For $n\ge4,$ prove that $F_n+1$ is not prime, where $F_n$ is $n^{th}$ Fibonacci number

What is the idea of the proof? I tried it by contradiction by letting $(1+F_n)$ to be prime $\implies$ $F_n$ is not prime $\implies$ WHAT NEXT?

Answer 1

As observed by Will Jagy we can always write $F_n+1$ as a product of a Fibonacci number and a Lucas number.

Let $\phi=(1+\sqrt5)/2$. It is well known that then $$ F_n=\frac1{\sqrt5}(\phi^n-(-\phi)^{-n}) $$ and $$ L_n=(\phi^n+(-\phi)^{-n}). $$ Both these sequences of integers satisfy the famous two-step recurrence relation, the difference coming from the initializations $F_0=0, F_1=1$ as opposed to $L_0=2, L_1=1$.

The following factorizations then follow immediately from $F_1=F_2=1$. $$ \begin{aligned} F_{2k+1}L_{2k}&=F_{4k+1}+F_1=F_{4k+1}+1,\\ F_{2k-1}L_{2k}&=F_{4k-1}+F_1=F_{4k-1}+1,\\ F_{2k}L_{2k-2}&=F_{4k-2}+F_2=F_{4k-2}+1,\\ F_{2k-1}L_{2k+1}&=F_{4k}+F_2=F_{4k}+1. \end{aligned} $$ All the residue classes modulo $4$ were covered, so the claim follows.

---

As an example: $$ \begin{aligned} F_{2k-1}L_{2k+1}&=\frac1{\sqrt5}(\phi^{2k-1}+\phi^{-(2k-1)})(\phi^{2k+1}-\phi^{-(2k+1)})\\ &=\frac1{\sqrt5}(\phi^{4k}+\phi^2-\phi^{-2}-\phi^{-4k})\\ &=F_{4k}+F_2=F_{4k}+1. \end{aligned} $$ It's all about polynomials of $\phi$. You do need to be careful with the parities of the exponents due to that $-\phi$ in the base.

Answer 2

Citing a proof of $F_n \pm 1$ not being a prime by tastymath75025 from AoPS forum: https://artofproblemsolving.com/community/c4h1249887p8924937

Lemma: If $n$ is odd, then $F_n^2-1=F_{n-1}F_{n+1}$.

Lemma 2: If $n$ is even, then $F_n^2-1=F_{n-2}F_{n+2}$.

Proof: For each statement either induct on $n$ or just use Binet's formula.

Now, if $n$ is odd then $(F_n-1)(F_n+1)=F_{n-1}F_{n+1}$. Clearly $F_n-1$ is not prime because $F_n-1 > F_{n-1}$ and $F_n-1 < F_{n+1} < 2(F_n-1)$, so $F_n-1$ cannot divide either factor on the RHS. Similar reasoning finishes for $F_n+1$.

Now, if $n$ is even then $(F_n-1)(F_n+1)=F_{n-2}F_{n+2}$. If $F_n-1$ is prime then clearly it must divide $F_{n+2}$ and not $F_{n-2}$. But it's easy to show $2(F_n-1) < F_{n+2} < 3(F_n-1)$, contradiction, and similarly for $F_n+1$.

As @lhf pointed out in comments, first two lemmas are known as Cassini and Catalan's identities. Also worth adding that $F_{n+1} < 2(F_n-1)$ is true for $n\geq 6$, while $F_{n+2} < 3(F_n-1)$ for $n\geq 7$. Rest of the cases can be checked by hand.

Answer 3

Apparently one and only one way to factor as a Lucas times a Fibonacci number.

Sun May 13 13:02:30 PDT 2018

4  1 + F 6 = 3  *  2  Lucas index 2 Fibonacci index  2
5  1 + F 9 = 3  *  3  Lucas index 2 Fibonacci index  3
6  1 + F 14 = 7  *  2  Lucas index 4 Fibonacci index  2
7  1 + F 22 = 11  *  2  Lucas index 5 Fibonacci index  2

8  1 + F 35 = 7  *  5  Lucas index 4 Fibonacci index  4
9  1 + F 56 = 7  *  8  Lucas index 4 Fibonacci index  5
10  1 + F 90 = 18  *  5  Lucas index 6 Fibonacci index  4
11  1 + F 145 = 29  *  5  Lucas index 7 Fibonacci index  4

12  1 + F 234 = 18  *  13  Lucas index 6 Fibonacci index  6
13  1 + F 378 = 18  *  21  Lucas index 6 Fibonacci index  7
14  1 + F 611 = 47  *  13  Lucas index 8 Fibonacci index  6
15  1 + F 988 = 76  *  13  Lucas index 9 Fibonacci index  6

16  1 + F 1598 = 47  *  34  Lucas index 8 Fibonacci index  8
17  1 + F 2585 = 47  *  55  Lucas index 8 Fibonacci index  9
18  1 + F 4182 = 123  *  34  Lucas index 10 Fibonacci index  8
19  1 + F 6766 = 199  *  34  Lucas index 11 Fibonacci index  8

20  1 + F 10947 = 123  *  89  Lucas index 10 Fibonacci index  10
21  1 + F 17712 = 123  *  144  Lucas index 10 Fibonacci index  11
22  1 + F 28658 = 322  *  89  Lucas index 12 Fibonacci index  10
23  1 + F 46369 = 521  *  89  Lucas index 13 Fibonacci index  10

24  1 + F 75026 = 322  *  233  Lucas index 12 Fibonacci index  12
25  1 + F 121394 = 322  *  377  Lucas index 12 Fibonacci index  13
26  1 + F 196419 = 843  *  233  Lucas index 14 Fibonacci index  12
27  1 + F 317812 = 1364  *  233  Lucas index 15 Fibonacci index  12

28  1 + F 514230 = 843  *  610  Lucas index 14 Fibonacci index  14
29  1 + F 832041 = 843  *  987  Lucas index 14 Fibonacci index  15
30  1 + F 1346270 = 2207  *  610  Lucas index 16 Fibonacci index  14
31  1 + F 2178310 = 3571  *  610  Lucas index 17 Fibonacci index  14

32  1 + F 3524579 = 2207  *  1597  Lucas index 16 Fibonacci index  16
33  1 + F 5702888 = 2207  *  2584  Lucas index 16 Fibonacci index  17
34  1 + F 9227466 = 5778  *  1597  Lucas index 18 Fibonacci index  16
35  1 + F 14930353 = 9349  *  1597  Lucas index 19 Fibonacci index  16

36  1 + F 24157818 = 5778  *  4181  Lucas index 18 Fibonacci index  18
37  1 + F 39088170 = 5778  *  6765  Lucas index 18 Fibonacci index  19
38  1 + F 63245987 = 15127  *  4181  Lucas index 20 Fibonacci index  18
39  1 + F 102334156 = 24476  *  4181  Lucas index 21 Fibonacci index  18

40  1 + F 165580142 = 15127  *  10946  Lucas index 20 Fibonacci index  20
41  1 + F 267914297 = 15127  *  17711  Lucas index 20 Fibonacci index  21
42  1 + F 433494438 = 39603  *  10946  Lucas index 22 Fibonacci index  20
43  1 + F 701408734 = 64079  *  10946  Lucas index 23 Fibonacci index  20

44  1 + F 1134903171 = 39603  *  28657  Lucas index 22 Fibonacci index  22
Sun May 13 13:02:30 PDT 2018

Answer 4

For all $n=3k,3k+1$, $F_n$ will be odd. Which means $F_n+1$ will be even
Still trying for $n=3k+2$ when $F_n$ will be even.

PS: $F_1=1$