-2

I'm investigating the Fibonacci sequence $F_{6n+2}+2$. I searched, by Maple, the first $10000$ numbers. I couldn't find any squares. I tried using quadratic residue and modularity but I got nothing worth investigating.

Is there a straightforward way to prove this?
Are there methods I'm not aware of? Any mention of a paper related to the problem would be appreciated.

GSmith
  • 144
ThePirateKing
  • 306
  • 1
    You should include your efforts. – jjagmath Jun 21 '24 at 11:38
  • Did you really try using modular arithmetic? – jjagmath Jun 21 '24 at 11:41
  • I tried using quadratic residue, but each time I get a subsequence that 'can be' a square. The process get repeating and no direct result is being deduced. I'm not searching for the answer, it could be a Paper Itself. Yet I want to know if there's a paper that could help. – ThePirateKing Jun 21 '24 at 11:44
  • There are only three square numbers in the Fibonacci sequences, see for example https://math.stackexchange.com/q/1012999/42969 – Martin R Jun 21 '24 at 11:47
  • Thanks for the paper Martin. As expected it's not that easy. I'm going to see If i could be inspired by this Paper. Thanks again. – ThePirateKing Jun 21 '24 at 11:50
  • 7
    It's very easy to see that $F_{6n+2}+2 \equiv 3 \pmod 4$, so it's not a square. – jjagmath Jun 21 '24 at 12:08
  • Thanks you so much. I know exactly what I've done wrong. I was only trying only $\mod p$ where $p$ is prime. I'm never going to forget you and the error i've made. – ThePirateKing Jun 21 '24 at 12:26
  • It even seems to be the case that $F_n+2$ is no perfect power for any non-negative integer $n$. – Peter Jun 23 '24 at 10:30
  • @Peter $F_3+2$ and $F_9+2$ say hello. Probably the only ones. – Oscar Lanzi Jun 26 '24 at 18:33
  • 1
    I apologize for this mistake. I must have mixed something up in the code. Thank you Oscar for pointing it out. – Peter Jun 26 '24 at 21:51

1 Answers1

5

Render the Fibonacci Sequence $\bmod 4$:

$1,1,2,3,1,0,1,1,2,...$

The sequence has period $6$ and $F_{6n+2}\equiv1\bmod4$. Therefore $F_{6n+2}+2\equiv3\bmod4$ and ... .

Oscar Lanzi
  • 48,208