TITLE: Remainders of Fibonacci numbers
QUESTION [13 upvotes]: Let $a>b$ be positive integers. Is there a Fibonacci number that is $b$ modulo $a$?
We know that the Fibonacci numbers are periodic modulo $a$. Indeed, consider pairs  $(F_i,F_{i+1})$ modulo $a$. There can only be finitely (around $a^2$) such pairs. So some two pairs coincide. Once we have that, we also have periodicity backward/forward. But it's still not clear for which $a,b$ we'll have $F_i\equiv b\pmod a$.

REPLY [2 votes]: There are infinitely counterexamples, and many can be found as follows. First, for a positive integer $a$, let $\kappa(a)$ denote the period of the Fibonacci numbers modulo $a$. That is, it is the smallest positive integer such that $F_{\kappa(a)+n}=F_n$ for all $n$. The number $\kappa(a)$ is known as the Pisano period, or the Wall number of $a$. As the values of the Fibonacci sequence modulo $a$ repeat after $\kappa(a)$, it can only have at most $\kappa(a)$ residues modulo $a$. So, as long as $\kappa(a) < a$ then there must exist $b < a$ such that no Fibonacci number is equal to $b$ mod $a$.
Actually, we can do a bit better as many of the residues modulo $a$ will be repeated within each period. We have,

$F_1=F_2=1$.
$F_{\kappa(a)-n}\equiv (-1)^{n-1}F_n$.

The first of these reduces the number of residue classes of the Fibonacci sequence by 1, and the second reduces it by the number of odd $n < \kappa(a)/2$, and there are $\lceil\kappa(a)/4-1/2\rceil$ possible values of $n$. Also, as long as $\kappa(a) > 3$, these relations are all independent, so there are at least
\begin{align}
a-\kappa(a)+1+\lceil\kappa(a)/4-1/2\rceil=\lceil a-3\kappa(a)/4+1/2\rceil&&{\rm(1)}
\end{align}
values of $b < a$ such that no Fibonacci number is equal to $b$ modulo $a$.
Note also that $F_3=2\not\equiv0$ for any $a > 2$, so we have $\kappa(a) > 3$ for all $a > 2$. So, there exists at least one $b < a$ such that no Fibonacci number is equal to $b$ modulo $a$ so long as $\kappa(a) < (4a+2)/3$.
To quickly compute upper bounds for $\kappa(a)$ the following facts can be used (summarised in this paper by Elsenhans and Jahnel).

$\kappa(2)=3$, $\kappa(5)=20$.
For prime $p\equiv\pm1$ (mod 5), then $\kappa(p)\mid(p-1)$
For odd primes $p\equiv\pm2$ (mod 5), then $\kappa(p)\mid2(p+1)$
For prime $p$ and $r\ge1$ then, $\kappa(p^r)\mid\kappa(p)p^{r-1}$.
If $a=a_1a_2\cdots a_N$ where $a_i$ are pairwise coprime, then $\kappa(a)$ divides the least common multiple of $\kappa(a_1),\kappa(a_2),\ldots,\kappa(a_N)$.

Just the second statement above is enough to generate infinitely many examples. For any prime $p\equiv\pm1$ (mod 5), we have $\kappa(p)\le p-1$, so there exists at least $(p+5)/4$ values of $b < p$ such that no Fibonacci number is equal to $b$ mod $p$. For example, taking $p=11$, a quick calculation shows that this is the case with $b=4,6,7,9$.
We can also find examples by using statement 5 above and multiplying together comprime $a_i$ where the $\kappa(a_i)$ have plenty of common factors.