9
$\begingroup$

It is well known that the last Fibonacci number $F_k$ such that $\exists \ n \in \Bbb{N} : F_k = n^2$ is $144$. Thus there are only $4$ perfect squares among the Fibonacci sequence (assuming you count $F_0 = 0$ and count the duplicated $F_1 = F_2 = 1$).

The equation $$F_k = n^2 + 4 $$ also has a few solutions ($F_5 = 1^2 + 4, F_6 = 2^2 +4, f_7 = 3^2+4$) and appears to have no other solutions (although my proof of that might well have some holes).

Pretty Hard Question

Prove that $$F_k = n^2 + 1 $$ and $$F_k = n^2 -1 $$ each have only a finite number of solutions for $k,n \in \Bbb{N}$.

Challenging Question

Are there any values of $s$ such that

$$F_k = n^2 +s $$ has no solutions for $k,n \in \Bbb{N}$?

Very Challenging Question:

Prove that for any $s \in \Bbb{Z}$ $$F_k = n^2 + s $$ has only a finite number of solutions for $k,n \in \Bbb{N}$.

(Or find a counterexample $s$.)

$\endgroup$
4
  • $\begingroup$ Even the result you said it's "well known" is not easy to prove $\endgroup$
    – Elaqqad
    Commented May 8, 2015 at 14:26
  • $\begingroup$ Yes, it is half as hard as my "pretty hard question." BUt you can find an understandable proof on the web. $\endgroup$
    – Mark Fischler
    Commented May 8, 2015 at 16:20
  • $\begingroup$ This website is not really a place to share hard problem but to help others in mathematics, If you're interested I have an old question here math.stackexchange.com/q/1149504/204937 $\endgroup$
    – Elaqqad
    Commented May 8, 2015 at 18:14
  • $\begingroup$ My suggestion would be to look at the $n^2$ proof, and see whether you can modify it to solve one/some of the other questions. $\endgroup$
    – Gerry Myerson
    Commented Nov 15, 2015 at 9:01

2 Answers 2

Reset to default
2
$\begingroup$

The pretty hard question is completely resolved in this paper. The authors show that the only perfect powers (not only perfect squares!) which are at distance $1$ from a Fibonacci number are $0,1,2,3,5,8$.

$\endgroup$
2
$\begingroup$

Pretty Hard Question: Partial Answer

$For\quad the\quad case\quad when\quad F_{ n }=m^{ 2 }+1,\quad F_{ n }-1=m^{ 2 }$

I. $n=4k\quad F_{4k}-1=F_{2k+1}L_{2k-1}$.

Note $L_{2k-1}=F_{2k+1}-F_{2k-3}$ where $gcd(F_{2k+1},F_{2k-3})=1$ Therefore, $L_{2k-1} and F_{2k+1}$ are relatively prime, and both must be squares. However, $L_{n}$ and $F_{n}$ both have only a finite number of solutions when they are square. Our proof is done.

II. $n=4k+1\quad F_{4k+1}-1=F_{2k}L_{2k+1}$.

Note $L_{2k+1}=F_{2k}+F_{2k+2}$ where $gcd(F_{2k+2},F_{2k})=1$. Therefore, $L_{2k+1} and F_{2k}$ are relatively prime, and both must be squares. However, $L_{n}$ and $F_{n}$ both have only a finite number of solutions when they are square. Our proof is done.

While I haven't been able to try the rest, I think they would follow a similar pattern.

Formula Used Presented in

Fibonacci[n]-1 is always composite for n>6. why?

https://en.wikipedia.org/wiki/Lucas_number

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .