We claim that the answer is yes. Take $m>>k$ such that $m=F_{2t}$ for some large integer $t$. Use Cassini identity $F_{2t}F_{2t+2}=F_{2t+1}^2-1$ to get that $F_{2t+1}^2 \equiv 1$ mod $F_{2t}$. Working mod $F_{2t}$ and using induction we can also see that $F_{2t+n} \equiv F_{n}F_{2t+1}$ mod $F_{2t}$ (base cases $F_{2t+1}, F_{2t+2}$ are easily verified and induction step is done using distributive property of multiplication and the linear recurrence). Since $F_{2t+1}^2 \equiv 1$ mod $F_{2t}$ by Cassini identity, we check that $F_{4t+n} \equiv F_{2t+n}F_{2t+1} \equiv F_{n}(F_{2t+1})^2 \equiv F_n$ mod $F_{2t}$. So we can see that the Fibonacci numbers give us at most $4t$ possible remainders mod $F_{2t}$. Also, using $F_{m}+F_{m+1}=F_{m+2}$ we can check that $F_{2t-1} \equiv F_{2t+1}$ mod $F_{2t}$ and $F_{2t-2} \equiv (-1)*F_{2t+2}$ mod $F_{2t}$. Using these as base cases we can apply induction and prove that for all non-negative integers $n \leq 2t$ we have $(-1)^{n+1}F_{2t-n} \equiv F_{2t+n}$ mod $F_{2n}$. Now using the earlier proved fact that $F_{4t+n} \equiv F_{n}$ mod $F_{2t}$, we can see that for any Fibonacci number $F_q$, it is congruent mod $F_{2t}$ to a number of the form $\pm F_v$ where $0 \leq v \leq 2t-1$. So now consider the positive integer $n$ for which $m=F_{2t}|F_n-k$. By the previous argument, $n$ is congruent to some number of the form $\pm F_r$ mod $F_{2t}$ where $0 \leq r \leq 2t-1$. Thus $F_{2t}|\pm F_r-k$ for some appropriate choice of sign. Now since $r$ is between $0$ and $2t-1$, we have $|F_r| \leq F_{2t-1}=F_{2t}-F_{2t-2}$. Now, since we chose $m=F_{2t}$ to be very large in comparison to $k$, we can see that $|k|<|F_{2t-2}|$. Thus $| \pm F_r-k|<F_{2t}$ and $F_{2t}| \pm F_r-k$, so we can see that $\pm F_r-k$ is forced to be equal to $0$. So $k=\pm F_r$ for some Fibonacci number less than $F_{2t}$. But $k$ is a positive integer and $0 \leq r \leq 2t-1$. Thus $k=F_r$. So $k$ has to be a Fibonacci number. Proved

For the sake of contradiction assume that there exists such $k$ which is not a Fibonacci number Then for all primes $p$ and some $m$ we have $F_{m}\equiv k\pmod{p}$ Note that $n$ is a Fibonacci number if and only if $5n^2-4$ or $5n^2+4$ is a perfect square. So for all primes either $5k^2-4$ or $5k^2+4$ is a quadratic residue modulo primes which is obviously a contradiction as for any two fixed integers we can have a prime such that both integers are quadratic non residue modulo prime

MarkBcc168 wrote: Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer m there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number? Proposed by Fedir Yudin. Solution. Yes, we claim that $k$ must be a Fibonacci Number. Define $g(n)$ to be least possible natural number satisfying $F_n\mid F_{g(n)}-k$ and set $F_0=0.$ Let $\phi=\tfrac{1+\sqrt 5}{2}.$ The following facts are well known and easy to prove $F_n=\left[\tfrac{\varphi^n}{\sqrt 5}\right]$ where $[\cdot]$ is the nearest integer function. $F_n=\tfrac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}.$ $F_{2n}>\varphi F_{2n-1}.$ $F_{m+n}=F_{m-1}F_n+F_mF_{n+1}.$ $F_{n-1}F_{n+1}-F_n^2=(-1)^n.$ $\gcd(F_m,F_n)=F_{\gcd(m,n)}.$ Claim 1. $F_{2n}\mid F_{4n}$ and $F_{2n}\mid F_{4n+1}-1.$ Proof. The first divisibility is trivial by (vi). For the latter part notice that (iv) implies $F_{4n+1}=F_{2n-1}F_{2n+1}+F_{2n}F_{2n+2}\equiv F_{2n-1}F_{2n+1}\pmod{F_{2n}}.$ Using (v) we have that $F_{2n-1}F_{2n+1}-F_{2n}^2=1.$ Hence the claim. $~\square$ The above claim implies that $g(2n)<4n.$ Assume that $k$ is not a Fibonacci number, so let us set $g(2n)=2n+p$ for now where $p\ge 1$. Using (iv) we have $k\equiv F_{2n+p}\equiv F_pF_{n-1}\pmod {F_n}.$ Let $a=\left\lfloor \tfrac{F_p}{\varphi}\right\rfloor$ and $b=F_p-a\varphi.$ Claim 2 $$|F_{n}-\varphi F_{n-1}|=\frac{\varphi^{-n}(2+\varphi)}{\sqrt 5}.$$Proof. \begin{align*} |F_{n}-\varphi F_{n-1}|&=\left|\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}-\varphi\left(\frac{\varphi^{n-1}-(-\varphi)^{-n+1}}{\sqrt 5}\right)\right|\\ &=\frac{1}{\sqrt 5}\left|\varphi^n-(-\varphi)^{-n}-\varphi^n+(-1)^{n-1}\varphi^{-n+2}\right|\\ &=\frac{1}{\sqrt 5}\left|(-\varphi)^{-n}+(-1)^{n}\varphi^{-n+2}\right|\\ &=\frac{1}{\sqrt 5}\left|\varphi^{-n}+\varphi^{-n+2}\right|\\ &=\frac{\varphi^{-n}(2+\varphi)}{\sqrt 5}.\tag*{$\square$} \end{align*}Claim 3. Let $R$ be the remainder when $F_pF_{n-1}$ is divided by $F_n$ where $p\le n-1.$ Then $$-1+\left\{\frac{F_p}{\varphi}\right\}F_{n-1}<R<1+\left\{\frac{F_p}{\varphi}\right\}F_{n-1}$$for all large $n$ where $\{\cdot\}$ denotes the fractional part. Proof. Let $b=\left\{\frac{F_p}{\varphi}\right\}$ and $a=\left\lfloor\frac{F_p}{\varphi}\right\rfloor.$ Note that $\lim_{n\to \infty} \frac{a}{\varphi^{p-1}}=5^{-\tfrac12}.$ Now $$F_pF_{n-1}-aF_n=a(\varphi F_{n-1}-F_n)+bF_{n-1}$$By claim 1 it is not hard to see that the claim follows. $\square$ Therefore it follows that $$-1+\left\{\frac{F_p}{\varphi}\right\}F_{2n-1}<k<1+\left\{\frac{F_p}{\varphi}\right\}F_{2n-1}\qquad(\star)$$for all large $n$ where $1\le p\le 2n-1$ Again let $b=\left\{\frac{F_p}{\varphi}\right\}$ and $a=\left\lfloor\frac{F_p}{\varphi}\right\rfloor.$ We have $F_p=a\varphi+b$ as usual. By claim 1 it follows that $F_{p-1}=\left[\frac{F_p}{\varphi}\right]$ hence if $p$ is odd then $a=F_{p-1}-1$ and otherwise $a=F_{p-1}.$ Therefore $$b=F_p-a\varphi= \begin{cases}F_p-F_{p-1}\varphi &,\text{ if }p\text{ is even}\\ F_p-F_{p-1}\varphi+\varphi &,\text{ otherwise }\end{cases}.$$If $p$ is odd, $F_p>F_{p-1}\varphi$ hence $$bF_{2n-1}=(\varphi+F_p-\varphi F_{p-1})F_{2n-1}=F_{2n-1}\cdot\frac{2+\varphi}{\varphi^p\sqrt 5}+F_{2n-1}\varphi\to+\infty$$as $n\to\infty.$ Therefore $p$ must be even for all large $n.$ In this case we obtain $$bF_{2n-1}=(F_p-\varphi F_{p-1})F_{2n-1}=F_{2n-1}\cdot\frac{2+\varphi}{\varphi^p\sqrt 5}$$Using the fact that $F_{2n-1}\approx \frac{\varphi^{2n-1}}{\sqrt 5}\implies bF_{2n-1}\approx \varphi^{2n-1-p}\cdot\frac{2+\varphi}5$ we can conclude that $2n-p$ is eventually constant. So let $p=2n-c$ for all large $n$ where $c$ is a constant natural number. Since $p$ is even hence $c$ is also even. Now we have that $F_{2n}\mid F_{2n-c}F_{2n-1}-k$ for all large $n.$ Notice that $F_{2n-c}\equiv (-1)^cF_cF_{2n-2}\pmod{F_n}\equiv F_cF_{2n-2}\pmod{F_{2n}}\equiv -F_cF_{2n-1}\pmod{F_n}.$ This implies that $F_{2n}\mid F_c+k$ since $F_{2n-1}^2\equiv F_{2n-1}F_{2n+1}\equiv 1\pmod{F_{2n}}.$ Taking $n\to+\infty$ we obtain a contradiction. And we are done.$~\blacksquare$

The answer is yes, and suppose for the sake of contradiction that such a $k$ exists, where $k$ is not a Fibonacci number. Then take $m = F_{2a}$ large enough so that $F_{2a-1} > k$. Let $S_i = F_i \mod F_{2a}$. Then $S_i = F_i$ for $1 \leq i < 2a, S_{2a} = 0$, and it is easy to show by induction that for $ 0 < i < 2a,$ $$ S_{2a+i} = \begin{cases} F_{2a-i} &,\text{ if }i\text{ is odd}\\ F_{2a} - F_{2a-i} &,\text{ if }i\text{ is even}\end{cases}$$Then we find $S_{4a+1} = S_{4a+2} = 1$, so $S_i$ repeats every $4a$ and thus produces no new values after $S_{4a}$. However since $F_n \equiv k \mod F_{2a}$ for some $n$ yet $k$ is not a Fibonacci number and $k < F_{2a-1}$, we have that $k$ must be of the form $F_{2a} - F_{2x}$ for some $1 \leq x \leq a-1$. However the least number of this form is $F_{2a} - F_{2(a-1)} = F_{2a-1}$, so $k > F_{2a-1}$, contradiction, showing no such $k$ exists. $\blacksquare$

The answer is that k must be a Fibonacci number. Claim 1: For all positive integers $m, n$ with $n\geq m\geq 2$, $F_n F_m=F_{n+1}F_{m-1}+F_{n+1-m}(-1)^{m-1}$ Proof: Note $F_nF_2=F_n \cdot\ 1=F_n$ and $F_{n+1}F_{1}+F_{n-1}(-1)^1=F_{n+1}-F_{n-1}=F_{n}$ so this is true for $m=2$ and $$F_{n}F_{3}=2F_{n}, F_{n+1}F_{2}+F_{n-2}(-1)^2=F_{n+1}+F_{n-2}=F_{n}+F_{n-1}+F_{n-2}=2F_{n}$$so this is also true for $m=3$. If it is true for $m-1$ and for $m$, then $$F_{n}F_{m+1}=F_{n}F_{m}+F_{n}F_{m-1}=F_{n+1}F_{m-1}+F_{n+1-m}(-1)^{m-1}+F_{n+1}F_{m-2}+F_{n+2-m}(-1)^{m-2}=F_{n+1}F_{m}+(-1)^{m-2}(F_{n+2-m}-F_{n+1-m})=F_{n+1}F_{m}+(-1)^mF_{n-m}$$so it is also true for $m+1$. By induction, this completes the proof of the claim. Claim 2: When taking the Fibonacci sequence modulo F_{n+1}, all residues R with $F_{n-1}<R<F_{n}$ are missing for $n\geq 5$ (this just ensures that the set of such R is nonempty) Proof: By letting $m=n$ in the above claim, $F_{n}^2=F_{n+1}F_{n-1}+F_{1}(-1)^{n-1} \Rightarrow F_{n+1}F_{n-1}-F_{n}^2=(-1)^n$, so $F_{n}^2 \equiv 1$ or $-1$ mod $F_{n+1}$. If $F_{n}^2 \equiv 1$ mod $F_{n+1}$, then when writing the sequence modulo $F_{n+1}$, we will get $$1, 1, 2, 3, ..., F_{n-1}, F_{n}, 0, F_{n}\times 1, F_{n}\times 1, F_{n}\times 2, ..., F_{n}F_{n-1}, F_{n}^2 (\equiv 1), 0, 1, 1, 2, 3... $$after which we see it will just repeat. If If $F_{n}^2 \equiv -1$ mod $F_{n+1}$, then when writing the sequence modulo $F_{n+1}$, we will get $$1, 1, 2, 3, ..., F_{n-1}, F_{n}, 0, F_{n}\times 1, F_{n}\times 1, F_{n}\times 2, ..., F_{n}F_{n-1}, F_{n}^2 (\equiv -1), 0, -1, -1, -2, -3, ..., -F_{n-1}, -F_{n}, 0, -F_{n}\times 1, -F_{n}\times 1,- F_{n}\times 2, ...,- F_{n}F_{n-1}, -F_{n}^2 (\equiv 1), 0, 1, 1, 2, 3 ... $$and we see that it will just repeat. Therefore the possible residues we can get are $$0, \pm F_{1}, \pm F_{2},..., \pm F_{n-1}, \pm F_{n}, \pm F_{n}F_{1}, \pm F_{n}F_{2}, ... , \pm F_{n}F_{n-1}, \pm F_{n}^2$$. Note $F_{n-1}<R<F_{n} \Leftrightarrow F_{n}>F_{n+1}-R>F_{n-1}$, so as none of $F_{1}, F_{2}, ... , F_{n-1}, F_{n}$ lie in the range $F_{n-1}<R<F_{n}$, this means that when taken modulo $F_{n+1}$, none of $-F_{1}, -F_{2}, ... , -F_{n-1}, -F_{n}$ do either. By Claim 1, we know $F_n F_m=F_{n+1}F_{m-1}+F_{n+1-m}(-1)^{m-1} \equiv (-1)^{m-1}F_{n+1-m}$ mod $F_{n+1}$ for $n\geq m\geq 2$. Therefore we get that $\pm F_{n}F_{1} \equiv \pm F_{n}$ mod $F_{n+1}$, and using the above, this means that $\pm F_{n}F_{1}, \pm F_{n}F_{2}, ... , \pm F_{n}F_{n-1}, \pm F_{n}^2$ in some order are congruent to $\pm F_{1}, \pm F_{2},..., \pm F_{n-1}, \pm F_{n}$ mod $F_{n+1}$ and we know that none of these leave a residue R in the required range. This completes the proof of the claim. If k is not a Fibonacci number, then let $F_{r-1}<k<F_{r}$. Then by the result of Claim 2, there is no positive integer n such that $F_{r+1} \mid F_{n}-k$ and so k must be a Fibonacci number.

Consider modulo $F_{2l - 2} + F_{2l}$ first $4l - 2$ elements of the sequence: $F_1, F_2, ..., F_{2l - 1}, - F_{2l - 2}, F_{2l - 3}, - F_{2l - 4}, F_{2l - 5}, ..., - F_{2i}, F_{2i - 1}, ..., - F_2, F_1, 0,$ and after this remainders will be the same. So, assume on the contrary, that $k$ isn't a Fibonacci number, picking $m = F_{2l - 2} + F_{2l} > k$ we must have that $k$ is of the form $F_{2l - 2} + F_{2l} - F_{2j}$, where $j \le l - 2$, but all these numbers are from the interval $[F_{2l} + F_{2l - 3} ; F_{2l} + F_{2l - 2}]$, but choosing $l$ large enough, it's obvious, that $k$ can't belong to simultaneously all intervals $[F_{2l} + F_{2l - 3} ; F_{2l} + F_{2l - 2}]$, where $l$ can be large enough, which concludes the proof.

mathfun5 wrote: The answer is yes, and suppose for the sake of contradiction that such a $k$ exists, where $k$ is not a Fibonacci number. Then take $m = F_{2a}$ large enough so that $F_{2a-1} > k$. Let $S_i = F_i \mod F_{2a}$. Then $S_i = F_i$ for $1 \leq i < 2a, S_{2a} = 0$, and it is easy to show by induction that for $ 0 < i < 2a,$ $$ S_{2a+i} = \begin{cases} F_{2a-i} &,\text{ if }i\text{ is odd}\\ F_{2a} - F_{2a-i} &,\text{ if }i\text{ is even}\end{cases}$$Then we find $S_{4a+1} = S_{4a+2} = 1$, so $S_i$ repeats every $4a$ and thus produces no new values after $S_{4a}$. However since $F_n \equiv k \mod F_{2a}$ for some $n$ yet $k$ is not a Fibonacci number and $k < F_{2a-1}$, we have that $k$ must be of the form $F_{2a} - F_{2x}$ for some $1 \leq x \leq a-1$. However the least number of this form is $F_{2a} - F_{2(a-1)} = F_{2a-1}$, so $k > F_{2a-1}$, contradiction, showing no such $k$ exists. $\blacksquare$ This is the best solution here

set $k=$ $\prod_{} p{_i} ^{r_i}$ put $m=k \implies f_u = \prod_{} p_{i} ^ {a_i} \times S$ such that $gcd(S,k)=1$ and $a_i \geq r_i$ put $m= \prod_{} p{_i}^{r_{i}+1} \times S$ now $ f_v = \prod p_{i}^{r_i}\times T$ and $gcd (T,Sk)=1$ BUT $f_{gcd(v,u)} $ $ = $ $gcd(f_{v},f_{u})$ = $k$ $\blacksquare$

$\mathrm{ }$

If $F_{a-2}<k<F_{a-1}$, there is a contradiction at $m=F_a$.

The answer is yes. We are given that for every $m\in \mathbb{N}$, there exists a Fibonacci number that is $k\pmod m$, with $k$ fixed. Consider the modulus $m=F_{2N}+F_{2N-2}$ for some $N$. All equivalences that follow will be in mod $m$. Claim: $F_{2N+a} \equiv (-1)^{a+1}F_{2N-a-2}$ for all $a\ge 0$. Proof: This is easy to see by induction, base case $a=0$ obvious. We have \begin{align*} F_{2N+a+2}&=F_{2N+a+1}+F_{2N+a} \\ &\equiv (-1)^{a+2} F_{2N-a-3} + (-1)^{a+1} F_{2N-a-2} \\ &\equiv (-1)^a [F_{2N-a-3} - F_{2N-a-2}] \\ &\equiv (-1)^{a+3} F_{2N-a-4}. \end{align*}This completes the induction. $\square$ Now, the residues of $(F_0,\ldots,F_{4N-3})$ mod $m$ are \[ (F_0, F_1,\ldots, F_{2N-1}, -F_{2N-2}, F_{2N-3}, -F_{2N-4}, F_{2N-5},\ldots, -F_2, F_1). \]Therefore, $F_{4N-2}\equiv F_0$, $F_{4N-1} \equiv F_1$. This means that the sequence $(F_i \mod m)$ for $i\ge 0$ is periodic with period $4N-2$. So actually all the residues possible are in the list above. Now, just take $N$ large enough so that $k+F_{2N-2} < m$, i.e. $k<F_{2N}$. Then, we can never have $k\equiv -F_{2N-2\ell}$ for any $\ell=1,\ldots,N-1$ since $k+F_{2N-2\ell} < m=F_{2N}+F_{2N-2}$. This forces $k\equiv F_\ell$ for some $0\le \ell \le 2N-1$. But since $F_{\ell},k < F_{2N} < m$, this actually forces $k=F_{\ell}$. The end.

MarkBcc168 wrote: Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer m there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number? Proposed by Fedir Yudin. We claim that the answer is $k = F_t$ for some $t \in \mathbb{N}$. Now, $a$ is a Fibonacci number if and only if either amongst $5a^2 \pm 4$ is a perfect square. So, $F_{t^\prime} \equiv k \pmod{p}$ for any prime $p$ which yields a contradiction as we can choose a large enough prime for which $k$ is not a quadratic residue (a lemma which states that given a natural number q, there exists a prime p such that $x^2 \equiv q \pmod{p}$ has no solutions. Hence, it forces us to have that $k$ is a Fibonacci number. Edit : Some people PMed me so here's the final conclusion logic : We have that $k \equiv F_{t^\prime} \equiv 5b^2 \pm 4 \pmod{p}$. Now, we consider if $k \equiv 5b^2 + 4 \pmod{p}$. Clearly, $b^2 \equiv \frac{k-4}{5} \pmod{p}$ and we can clearly choose a sufficiently large enough prime $p$ for which the fraction $\frac{k-4}{5}$ is not a quadratic residue using aforementioned lemma unless $k = 4, 9$ for which the given claim can be contradicted too and similarly we continue for $k \equiv 5b^2 -4$.

babubhaiyya123 wrote: We claim that the answer is $k = F_t$ for some $t \in \mathbb{N}$. Now, $a$ is a Fibonacci number if and only if either amongst $5a^2 \pm 4$ is a perfect square. could you prove the second claim?

Eliot wrote: babubhaiyya123 wrote: We claim that the answer is $k = F_t$ for some $t \in \mathbb{N}$. Now, $a$ is a Fibonacci number if and only if either amongst $5a^2 \pm 4$ is a perfect square. could you prove the second claim? This follows from the method of solving Pell's equation.

Take the smallest $d>0$ such that $k|F_d$. Hence there is $n$ such that $$F_d|F_n-k$$So $k|F_n$. By $d$'s minimality, $d|n$ and therefore $F_d |F_n$, so $F_d|k$, hence $k=F_d$.

Gomes17 wrote: Take the smallest $d>0$ such that $k|F_d$. Hence there is $n$ such that $$F_d|F_n-k$$So $k|F_n$. By $d$'s minimality, $d|n$ and therefore $F_d |F_n$, so $F_d|k$, hence $k=F_d$. Wow this is nice

Take $m = F_s$ to be a fibonacci number divisible by $k^2$ Suppose $F_s \mid F_n - k$ We have $F_{\gcd(s,n)} = \gcd(F_n,F_s) = k$

We have $4$ lemmas from the problem $F_{nk} $ is a multiple of $F_k \quad (\star) $ $F_{(m+n)}=F_mF_{(n+1)}+F_nF_{(m-1)} \quad (\bullet) $ $\sum_{i=0}^{2n+1}(2n+1i)F_i^2=5^n\cdot F_{(2n+1)}\quad (\square)$ $\gcd(F_m,F_n)=F_{(\gcd(m,n))}\quad(\diamondsuit)$ We will prove Conjecture- For every prime $p,$ there exists a number $n$ such that $F_n=1\pmod{p}$ and $F_{n+1}=0\pmod{p}\implies F\,\text {is the periodic with the interval}=p. $ It can be solved by $(\bullet) $ and $(\square). $ This conjecture only $\implies (\square)$ when $p>5.$ We can verify that when $p=2$ $ F_{3\cdot 2^k}$ is divisible by $2^{k+1}.$ $p=3$ $F_{4\cdot 3^k}$ is divisible by $3^{k+1}$ $F_{5^k}$ is divisible by $5^k $

Conjecture 2- For prime $p$ and the number $k,$ if $n$ is the smallest number satisfies $F_n $ is divisible by $p\implies F_{n\cdot p^k}\,\text{is divisible by}\,p^{k+1}.$ This can be proved by induction, with the periodicity of $F$ in $\pmod{p}.$ From $(\star)$ and $(\diamondsuit)$ we can deduce that for every number $m,$ there exists index $n $ such that $F_n $ is the multiple of $m. $ There exists an index $n $ such that $p^k\mid F_n$ could be proved by applying pigeonhole theorem. Proof- Assume at step $k, $ there exists an index $n $ such that $p^k\mid F_n. $ Consider all the values of $F_{nq}.$ There always exists $2$ values $q_1$ and $q_2$ such that $F_{nq_1}=F_{nq_2}(\pmod{p^{k+1}}).$ Applying $(\bullet)$ note that all $F_{nq+1} $ or $F_{nq-1} $ has the remainder $=1\text {modulo p},$ we obtain the desired index $q $ such that $F_{nq}=0 (\pmod {p^{k+1}}). $ Back to the main problem For each number $k $ which is not a $\text{Fibonacci} $ number we take the smallest number $q$ such that $k\mid F_q.$ Choose $m=F_q. $ From this hypothesis $,F_n=k\pmod{F_q} $ We obtain $k\mid\gcd(F_n,F_q)=F_{\gcd(n,q)}$ combining with $q\implies q\mid\gcd(n,q)\implies q\mid n.$ So $F_q\mid k $ leads to a contradiction. For the converse part $, $ if $k=F_q. $ For every number $m, $ there exists $r $ such that $F_r$ is divided by $m.$ Choose number $n $ such that $n=q\pmod{r}\implies F_n=F_q\pmod{F_r}.$

Gomes17 wrote: Take the smallest $d>0$ such that $k|F_d$. Hence there is $n$ such that $$F_d|F_n-k$$So $k|F_n$. By $d$'s minimality, $d|n$ and therefore $F_d |F_n$, so $F_d|k$, hence $k=F_d$. how is it that nice

I think that all of these posts were quite similar though this one involved quadratic residues which was already mentioned though I mean there arent really that many solutions that you can think of, I edited this one a bit though.

The answer is yes. Lemma: For all $k\ge 2$, $F_k^2 + (-1)^k = F_{k-1}F_{k+1}$. Proof: We use induction. The base case is immediate. For $k\ge 3$, write \[F_{k-1}F_{k+1} = F_{k-1}^2 + F_{k-1}F_k = F_{k-2}F_k - (-1)^{k-1} + F_{k-1}F_k = F_k^2 + (-1)^k\]by the inductive hypothesis. This enables us to determine the residues of Fibonacci numbers modulo other Fibonacci numbers much more efficiently. In particular, note that as $F_k\equiv 0\pmod{F_k}$ and $F_{k+1} \equiv F_{k-1}\cdot 1\pmod{F_k}$, we have $F_a \equiv F_{k-1} \cdot F_{a-k}\pmod{F_k}$ for all $a>k$, and so $F_a\equiv F_{a-4k}\pmod{F_k}$ for all $a>4k$. Then for each $n$, to check whether it is congruent to a Fibonacci number modulo $F_k$ it is sufficient to check the Fibonacci numbers $F_1, F_2, \dots, F_{4k}$. It is well-known that every positive integer divides a Fibonacci number (use pigeonhole principle then work backwards), so we only need to worry about Fibonacci numbers anyway. In fact, every positive integer must divide a Fibonacci number of the form $F_{2k}$. So consider some positive integer $n$ that is not a Fibonacci number but is congruent to Fibonacci numbers modulo $F_{2k}$ and $F_{2k-2}$. Then by an analogous argument before, we can say $n\equiv F_{t_1}$ or $n\equiv F_{t_1}F_{2k-1}$ modulo $F_{2k}$ for some $1\le t_1\le 2k$, and $n\equiv F_{t_2}$ or $n\equiv F_{t_2}F_{2k-1}$ modulo $F_{2k-2}$ for some $1\le t_2\le 2k-2$. We check beforehand that $F_{2k-2} > n$. Thus it is certainly not congruent to $F_{t_1}$ or $F_{t_2}$ modulo either. Moreover, this ensures $t_2\ne 2k-2$ and $t_1\ne 2k$ because otherwise $n$ is a Fibonacci number, contradicting the assumption. Observe that $t_1 = 2k-1$ would yield $n\equiv F_{2k-1}^2 \equiv F_{2k}F_{2k-2} + 1 \equiv 1\pmod{F_{2k}}$. As $1$ is a Fibonacci number, this would also contradict the assumption. So $1\le t_1\le 2k-2$. Analogously, $1\le t_2\le 2k-4$. Let $n = F_{t_1}F_{2k-1} - aF_{2k}$ for some $a < F_{2k-1}$ and $n = F_{t_2}F_{2k-1} - bF_{2k-2}$ for some $b < F_{2k-1}$. To see the inequalities, note \[aF_{2k} < F_{t_1}F_{2k-1} < F_{2k}F_{2k-1}, bF_{2k-2} < F_{t_2}F_{2k-1} < F_{2k-2}F_{2k-1}.\]Then \[(F_{t_1} - F_{t_2})F_{2k-1} = aF_{2k} - bF_{2k-2} = aF_{2k-1} + (a-b)F_{2k-2}.\]Modulo $F_{2k-1}$, this implies $a=b$ because $\gcd(F_{2k-2},F_{2k-1}) = 1$ and $|a-b| < F_{2k-1}$. Moreover, $b = a=F_{t_1} - F_{t_2}$. So \[n = F_{t_1}F_{2k-1} - (F_{t_1} - F_{t_2})F_{2k} = F_{t_1}(F_{2k+1} - F_{2k}) - (F_{t_1} - F_{t_2})F_{2k} = F_{t_1}F_{2k+1} - (2F_{t_1}-F_{t_2})F_{2k}.\]But by an analogous argument, supposing $n$ is a Fibonacci number modulo $F_{2k+2}$ as well, we can also write this number as \[F_{t_4}F_{2k+1} - (F_{t_3} - F_{t_4})F_{2k}\]for some $t_3 \le 2k, t_4 \le 2k-2$. Thus, we have \[(F_{t_4} - F_{t_1})F_{2k+1} = F_{2k}\cdot (F_{t_3} - F_{t_4} + F_{t_2} - 2F_{t_1}).\]It is clear that $\gcd(F_{2k}, F_{2k+1}) = 1$, so because $F_{2k}\mid F_{t_4} - F_{t_1}$, we must have $t_4 = t_1$. Thus $F_{t_3} + F_{t_2} = 3F_{t_1}$. As $F_{t_2} < F_{t_1}$ and $F_{t_3} > F_{t_4} = F_{t_1}$, we must have $2F_{t_1} < F_{t_3} < 3F_{t_1}$. This can only occur if $F_{t_3} = F_{t_1+2}$. This is because $F_{t_1} < F_{t_1+1} < F_{t_1} + F_{t_1-1}$ and $F_{t_1+2} = F_{t_1} + F_{t_1} + F_{t_1-1}$. So $t_1 + 2 = t_3$ and $t_1 - 2 = t_2$. So the original formula was \[n = F_{t_1}F_{2k-1} - F_{t_1-1}F_{2k} = (F_{t_1-1}+F_{t_1-2})F_{2k-1} - F_{t_1-1}(F_{2k-1}+F_{2k-2}) = F_{t_1-2}F_{2k-1} - F_{t_1-1}F_{2k-2} = \]\[F_{t_1-2}(F_{2k-2}+F_{2k-3}) - (F_{t_1-2}+F_{t_1-3})F_{2k-2} = F_{t_1-2}F_{2k-3} - F_{t_1-3}F_{2k-2}.\]Iterating, we eventually get that for some $x$, either $n = 2F_{2x-1} - F_{2x} = F_{2x-1} - F_{2x-2} = F_{2x-3}$ or $n = F_{2x-1} - F_{2x} = -F_{2x-2}$. Either way, we get a contradiction, so we are done.

The answer is yes. Let $\alpha=\tfrac{1+\sqrt{5}}{2}$, so $F_n=\tfrac{\alpha^n-(-\alpha)^{-n}}{\sqrt{5}}$. It is well-known that $k$ is a Fibonacci number iff at least one of $5k^2-4$ and $5k^2+4$ is a perfect square. Suppose that $k$ was not a Fibonacci number, so neither $5k^2-4$ nor $5k^2+4$ are squares. Let $p_1,\ldots,p_a$ be the odd primes dividing $5k^2-4$ with $\nu_{p_i}(5k^2-4)$ odd, and $q_1,\ldots,q_b$ be the odd primes dividing $5k^2+4$ with $\nu_{q_i}(5k^2+4)$. Note that these are all distinct, since $\gcd(5k^2-4,5k^2+4) \mid 8$, and none of these can be $5$. If both $a \geq 1$ and $b \geq 1$, then by Dirichlet pick some prime $p \equiv 1 \pmod{8}$, $p \equiv 1 \pmod{5}$, $p \equiv 1 \pmod{p_i}$ and $p \equiv 1 \pmod{q_i}$ for $i \geq 2$, and $p$ equivalent to non-quadratic residues modulo $p_1$ and $q_1$. Because $p \equiv 1 \pmod{8}$, we have $(\tfrac{p}{q})(\tfrac{q}{p})=1$ for any odd prime $q$, and $(\tfrac{2}{p})=1$. Then, $$\left(\frac{5k^2-4}{p}\right)=\left(\frac{\varepsilon p_1\ldots p_a}{p}\right)=\left(\frac{\varepsilon}{p}\right)\left(\frac{p_1}{p}\right)\ldots\left(\frac{p_a}{p}\right)=\left(\frac{p}{p_1}\right)\ldots\left(\frac{p}{p_a}\right)=-1,$$where $\varepsilon \in \{1,2\}$ based on the parity of $\nu_2(5k^2-4)$. Likewise, $(\tfrac{5k^2+4}{p})=-1$. If $\min\{a,b\}=0$, i.e. one of $5k^2-4$ and $5k^2+4$ is twice a perfect square, then the another one cannot be by looking at $\nu_2$, so WLOG let $a=0$ and $b \geq 1$. Then pick some prime $p \equiv 5 \pmod{8}$, $p \equiv 1 \pmod{5}$, $p \equiv 1 \pmod{q_i}$ for $i \geq 2$, and $p$ equivalent to a non-quadratic residue modulo $q_1$. We still have $(\tfrac{p}{q})(\tfrac{q}{p})=1$, but now $(\tfrac{2}{p})=-1$. For a similar reason to before, $(\tfrac{5k^2+4}{p})=-1$, and $(\tfrac{5k^2-4}{p})=(\tfrac{2}{p})=-1$. Furthermore, for this value of $p$, since $(\tfrac{5}{p})=(\tfrac{p}{5})=1$, $\sqrt{5} \in \mathbb{F}_p \implies \alpha \in \mathbb{F}_p \implies \alpha^n \in \mathbb{F}_p$. Therefore, one of the following equations must have a root in $\mathbb{F}_p$: $$x+\frac{1}{x}=k\sqrt{5} \iff x^2-k\sqrt{5}+1=0 \text{ or } x-\frac{1}{x}=k\sqrt{5} \iff x^2-k\sqrt{5}-1=0.$$By using the quadratic formula, this means that either $5k^2-4$ or $5k^2+4$ is a square in $\mathbb{F}_p$, but this contradicts the construction of $p$. $\blacksquare$