Nearly similar to my previous question, is there a Fibonacci Number that is also a Carmichael Number?
We all know that for $n$ to be a Carmichael Number, it must be square-free and odd.
$F_{n}$ cannot be a Carmichael Number if $n$ is:
$n\equiv0,6\pmod{12}$, as $F_{n}$ will be divisible by either $4$ or $9$, hence not a Carmichael Number.
If $n\equiv0,110\pmod{121}$, then it is not a Carmichael Number as it will be divisible by $11^2$.
For $n$ to be a Carmichael Number, $n$ must not be equal to $0,91\pmod{169}$, as if $n\equiv0,91\pmod{169}$, then $13^{2}\mid F_{n}$.
If $17\mid F_{n}$, then $n$ must not be equal to $0, 153\pmod{289}$.
If $F_{n}$ is divisible by $19$, then $n$ must not be equal to $0,342\pmod{361}$, as the number will be divisible by $19^2$.
This is my knowledge only. My instinct is that a Fibonacci Number will never be a Carmichael Number.
I Am correct?
