We present a variation of the official solution by using a lemma mentioned in solutions to the 2010 Romanian Masters in Math; see https://aops.com/community/c6h334682p1798256. Lemma: Let $P$ be a set of $k$ primes. An arithmetic progression $x_1 < x_2 < \dots$ of positive integers is given, such that no prime in $P$ divides all terms in the progression. Then there exists $1 \le i \le 2^k$ such that $x_i$ is not divisible by any prime in $P$. Assume the arithmetic progression is decreasing and write it as \[ \frac{a_1}{b_1} = \frac{a+(n-1)r}{d}, \quad \frac{a_2}{b_2} = \frac{a+(n-2)r}{d}, \quad \dots, \quad \frac{a_n}{b_n} = \frac{a}{d} \]for some positive integers $a$, $r$, $d$ with $\gcd(a,r,d) = 1$. Note that one can choose $d$ in such a way that $d \mid b_1 b_2$, hence $d \le (5n)^2$. The idea is to again find numerators not divisible by $d$, by applying the lemma to the set $P$ of primes dividing $d$. Let $k = |P|$. We consider two cases. Assume $d \le 5n$. We claim that $n > 36 \cdot 2^k$ in this case. Indeed, since $n \ge 2018$ we only consider $k \ge 6$ (since $36 \cdot 2^5 < 2018$). But then \[ d \ge 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17^{k-6} \ge 30030 \cdot 2^{k-6} > 5 \cdot 36 \cdot 64 \cdot 2^{k-6}= 5 \cdot 36 \cdot 2^k. \]Hence $n \ge \frac15d = 36 \cdot 2^k$, as claimed. We now derive a contradiction by noting that among the first $6 \cdot 2^k$ fractions, at least six of them have denominator divisible by $d$, since the numerators share no common factor with $d$. Hence one of the fractions, say the $i$th one, must have denominator at least $6d$. The corresponding numerator is then at least \[ 6[a + (n-i)r] \ge 6[a + (n-6\cdot2^k)r] > 6\left[a + \frac56 n \cdot r\right] > 5n \]which is a contradiction. Assume $d > 5n$, but still $d \le (5n)^2$. We claim that $n > 2^k$ in this case. Indeed, since $n \ge 2018$ we only consider $k \ge 11$ (since $2^{10} < 2018$). Then \[ d \ge 2 \cdot 3 \cdot 5 \cdot \dots \cdot 31 \cdot 37^{k-11} > 4^{11} \cdot 25 \cdot 4^{k-11} = 25 \cdot 4^{k}. \]Then $n \ge \sqrt{d/25} > 2^k$. But now among the first $2^k$ fractions, one of them is irreducible again, but then its denominator is a multiple of $d$, despite $d > 5n$. This is again a contradiction.

This problem is proposed by Pakawut Jiradilok and Warut Suksompong, Thailand, also my favorite problem in ISL (minus IMO). We begin with the following crucial $p$-adic lemma. Lemma: Let $a_1,a_2,...,a_n$ be an arithmetic progression of rational number and $k\in\mathbb{Z}^+$. Let $t=\min\{\nu_p(a_1),\nu_p(a_2),...,\nu_p(a_n)\}$. Then at most $\left\lceil\frac{n}{p^k}\right\rceil$ of them have $\nu_p$ greater of equal to $t+k$. Proof: By suitable scaling, WLOG $a_1,a_2,...,a_n$ are integer such that $\gcd(a_1,a_2,...,a_n)=1$. Then $t=0$. Let $a_i = a+ib$ and take $S=\{i : p^k\mid a_i\}$. Then if $i,j\in S$, then $$p^k\mid (a+ib) - (a+jb) = (i-j)b.$$If $p\mid b$ then $p\mid a$, contradiction to $t=0$. Thus $i,j\in S\implies p^k\mid i-j$. Thus $S$ is subset of some residue class $\pmod{p^k}$, implying the conclusion. Using our lemma, we give the key claim. Claim: For each prime power $q=p^k > 5$, we have $\nu_p(\frac{a_i}{b_i}) > -k$ for each $i=1,2,...,n$. Proof: Assume for the contradiction that $\min\{\nu_p(\frac{a_1}{b_1}), \nu_p(\frac{a_2}{b_2}),,...,\nu_p(\frac{a_n}{b_n})\} < -k$ then by our lemma, at most $\left\lceil\frac{n}{p^k}\right\rceil$ of all fractions have $\nu_p$ greater or equal to zero. If $\nu_p(\frac{a_i}{b_i})<0$, then $p\mid b_i$. As there are at most $\frac{5n}{p}$ multiples of $p$ in $\{b_1,...,b_n\}$, we get $$n - \frac{n}{q} - 1 \leqslant \frac{5n}{p} \leqslant \frac{5n}{q}$$, a contradiction for $q \geqslant 7$. The claim implies $60\cdot\frac{a_i}{b_i}\in\mathbb{Z}$ for each $i=1,2,...,n$. Let $x_i = \frac{60a_i}{b_i}$, we split into two cases. Case I. $\gcd(x_1,x_2,...,x_n)=1$} Then at least $\frac{4}{15}$ of $x_i$ are relatively prime to $60$, these terms force $b_i$ to be multiple of $60$. However, there are at most $\frac{5}{60}=\frac{1}{12}$ multiples of $60$, contradiction. Case II. $\gcd(x_1,x_2,...,x_n)>1$ Then $x_i\geqslant 2i$ for each $i$. Thus $$2\cdot 4\cdot 6\cdot...\cdot 2n = x_1x_2...x_n = \frac{60^n(a_1a_2...a_n)}{b_1b_2...b_n} \leqslant \frac{60^n(4n+1)(4n+2)...(5n)}{n!}$$or $(n!)^2(4n)!\leqslant 30^n(5n)!$. An integral estimation implies $(\frac{n}{e})^n < n! < (\frac{n+1}{e})^{n+1} < (\frac{n}{2.5})^n$ for $n\geqslant 2018$. After rearranging terms, we get $$\left(30\cdot 2^5\cdot \frac{e^6}{4^4}\right)^n \geqslant n^n \implies n \leqslant 30\cdot 2^5 \cdot \frac{e^6}{4^4} < 1920$$, contradiction. (The computation is resonably derived by noting that $e^6 < 512 \iff e^2 < 8$.)

Let $x_i = a_i/b_i$ and suppose $x_1 < x_2 < \cdots < x_n$. Let $d = x_2 - x_1$. We claim the following: Lemma: If $v_p(d) = -M$ for $M>0$, then 1. $\frac{2017}{2018} < \frac{5}{p^M} + \frac{1}{p}$ 2. If $v_p(x_k) > -M$ then $k\equiv r\pmod{p}$ for some fixed integer $r$. Proof: We do this by showing $p^M$ must divide at least $n - n/p-1$ of the $b_i$'s. We do this via casework on $v_p(x_1)$: 1. $v_p(x_1) < v_p(d)$. In this case, $v_p(x_k) = v_p(x_1 + (k-1)d)$. Since $v_p(x_1) < v_p((k-1)d)$, it follows $v_p(x_k) = v_p(x_1) < -M$ for all $k$, so $v_p(b_k) > M$ and thus $p^M$ divides all $n$ of the $b_i$'s. 2. $v_p(x_1) \ge v_p(d)$. This is equivalent to $v_p(p^Mx_k) > 0$, or $v_p(p^Mx_1 + (k-1)p^Md) > 0$. Now, note that $v_p(p^Md) = 0$ and $v_p(p^Mx_1)\ge 0$, so we can treat this as a modular expression: $(p^Mx_1) + (k-1) (p^Md)\equiv 0\pmod{p}$. Then it follows $k \equiv 1 - (p^Md)^{-1} (p^Mx_1)\pmod{p}$ where taking the modular inverse is defined since $v_p(p^Md) = 0$. Now, the number of integers between $1$ and $n$ that are equivalent to $k\pmod{p}$ is less than $n/p + 1$, so more than $n - n/p - 1$ values of $k$ do not satisfy $k\equiv r\pmod{p}$, implying more than $n-n/p - 1$ values of $k$ satisfy $v_p(x_k) = -M$. For these values of $k$, we have $v_p(b_k) \ge M$, so we have $p^M$ divides more than $n-n/p - 1$ of the $b_i$'s. On the other hand, the $b_i$'s are distinct, and there are at most $5n/p^M$ multiples of $p^M$ among them, so we must have $n-n/p-1<5n/p^M$ which implies $\frac{n-1}{n} < \frac{1}{p} + \frac{5}{p^M}$. Since $n \ge 2018$, we get the desired lemma. $\blacksquare$ Now, solving the inequality easily gives $p^M\in \{2, 4, 8, 3, 5\}$, so if $v_p(d) < 0$ then $p\in \{2, 3, 5\}$. Now, suppose $x_k$ and $d$ have the same denominator when simplified; that is, $v_p(x_k) = v_p(d)$ whenever $v_p(d) < 0$. Then by CRT and the lemma above, it follows that there are at least $8$ residues $r$ such that we can have $k\equiv r\pmod{30}$. Let $D$ be the denominator of $d$ when written in lowest terms. Then by the above discussion, it follows that $D\mid b_i$ for at least $\lfloor n/30\rfloor\cdot 8$ of the $b_i$'s. It thus remains to bound $D$. Now, clearly $d\ge \frac{1}{D}$, so $x_k > \frac{k-1}{D}$. Moreover, among the fractions $x_i$, note that $b_i\ge n/2$ for at least $n/2$ of the $b_i$'s. For these terms, note that $x_i = a_i/b_i \le (5n)/(n/2) = 10$, so it follows $x_{n/2} \le 10$. Then $\frac{n/2-1}{D} < 10$, so $n < 2(10D + 1) = 20D + 2$. Thus $n\le 20D+1$, so there are at most $5n/D \le 105$ multiples of $D$ between $1$ and $5n$. But $D\mid b_i$ for at least $\lfloor n/30\rfloor\cdot 8 < 8(n/30 - 1)$ of the $b_i$'s. Thus $8(n/30 - 1) \le 105$ which is an obvious contradiction, as desired. $\square$

In fact, we can generalize $5n$ to $n^{3/2-\epsilon}$ for $n$ large enough. Let $f(n) = n^{1/2-\epsilon}$; then, replacing $5$ with $f(n)$ in the problem, and adopting the notation above, we see that if $D$ is the denominator of common difference, then for $p\mid D$ we must have $\frac{n-1}{n} < \frac{1}{p} + \frac{f(n)}{p^M}$, which is equivalent to $p < f(n) + 1$. Next, by CRT, the fraction of $b_i$'s that are divisible by $D$ is at least $\prod_{p < f(n) + 1} \left(1-\frac{1}{p}\right) \approx \prod_{p < f(n) + 1} e^{-1/p} \approx \frac{1}{\ln (f(n) + 1)}$, so there must be at least $n/\ln (f(n) + 1)$ multiples of $D$ less than $n\cdot f(n)$. On the other hand, we get $n\le 4f(n)\cdot D+ 1$ so the number of multiples of $D$ less than $f(n)\cdot n$ is at most roughly $f(n)(4f(n) + 1)$. But $f(n)(4f(n) + 1) = O(n^{1-2\epsilon})$ while $n/\ln (f(n) + 1) = O(n/\ln n)$ which gives our contradiction.

For simplicity, let say the AP is $\frac{a_0}{b_0},...,\frac{a_{n-1}}{b_{n-1}}$ given by $\frac{a_k}{b_k}= \frac{u+kt}{v}$. We choose $u,v,t$ in such a way that $\gcd (u,v,t) =1$. Assume WLOG $t>0$ (if $t=0$ we are done and if $t<0$ we can reverse the sequence. ) Notice that since the denominators are distinct, $$\min( \frac{a_{n-1}}{b_{n-1}}, ... , \frac{a_{n-k}}{b_{n-k}}) \le \frac{5n}{k}$$ Hence $$\frac{(n-k)t}{v} < \frac{u+(n-k)t}{v} \le \frac{5n}{k}$$ By taking $k= \frac{n}{2}$ or $\frac{n-1}{2}$ if $n$ is odd, we get $$t<\frac{20nv}{n^2-1}$$ Now we prove that $v$ must be very small. Together with the last bound on $t$, we will get $t <1$, contradicting our hypothesis that $t \neq 0$. Suppose that a prime power $p^e$ divides $v$. Hence we must have that, since $p$ doesn't divide $\gcd (u,t)$, that at most one reminder $r \mod p$ gives $p|u+rt$. The conclusion is that $p^e$ must divide $b_i$ for all but, at most, $\lceil \frac{n}{p} \rceil < \frac{n}{p} +1$ values of $i$. Therefore, since all $b_i$ are distinct, we must have $(n- \frac{n}{p}-1)p^e \le 5n$ (here we are counting the number of multiples of $p^e$ less than $5n$). Its easy to verify that if $p \ge 7$ is any prime or $p^e \ge 9$, then the last inequality fails. Therefore $v \le 8 \times 3 \times 5 = 120$. We will prove that it can not happen that $3$ and $5$ divide $v$ simultaneously. If that happened, we can see that there are $4$ reminders $r \mod 5$ such that $5$ divides $b_i$ when $i \equiv r \mod 5$. With the same idea, there are two such reminders $\mod 3$. Hence, for at least each $8$ of every $15$ consecutive $i$, we have $15 | b_i$. Therefore $8 \lfloor \frac{n}{15} \rfloor \le \frac{5n}{15}$ (counting the multiples of $15$ less than $5n$), which is clearly not true when $n \ge 2018$. Now we see that $v \le 40$, hence $t < \frac{800n}{n^2-1} <1$, contradiction. Remark. Notice that this last step of proving that $3$ and $5$ couldn't divide $v$ at the same time was very important, since without this step we would have only proved our statement for $n \ge 2400$.

Nice problem! Solved with Pujnk. (There are possibly flaws in this proof, please do point out any if you spot them) Assume WLOG the sequence is increasing and let the common difference be $\frac{c}{d}$ with $c,d$ coprime with $d > 0$. So we have $\frac{a_i}{b_i} + (j-i)\frac{c}{d} = \frac{a_j}{b_j}$ $$\implies d(a_jb_i - a_ib_j) = (j-i)cb_ib_j$$Call this $(1)$. In particular, this means $d | (j-i)b_ib_j$ since $c,d$ coprime. The main idea of the problem is that many of the $b_i$ have to be divisible by primes dividing $d$ and since the $b_i$ cant be big, $d$ itself must be small. But if $d$ is small, then the sequence grows too fast to have only small $a_i, b_i$ Claim: $d \le 120$ Proof: Let $p$ be the largest prime dividing $d$. We have that $p | (j-i)b_ib_j$ for all $i,j$. Suppose $b_k$ is a term such that $p \nmid b_k$ and say $k \equiv r \pmod p$. For every index $i$ such that $p \nmid i-k$, we have that $p | b_i$. Since there are $p-1$ such indices, we have that among $b_1, b_2, \cdots b_n$, we have at least $\left \lfloor \frac{n(p-1)}{p} \right \rfloor$ terms divisible by $p$. Since we have $b_i \le 5n$, there are at most $\left \lfloor \frac{5n}{p} \right \rfloor$ numbers divisible by $p$ and $\le 5n$, (ignore floors because the inequality works anyway), $\frac{n(p-1)}{p} \le \frac{5n}{p} \implies p \le 6$. So, $d$ has only prime factors $2,3,5$ Next, we bound the highest power of these primes that can divide $d$. Say $d = 2^x 3^y 5^z$. Let $p$ be a prime and from $(1)$, we have that $v_p(d) \le \max(v_p(b_i), v_p(b_j)) + v_p(j-i)$ since there's at least a $v_p$ of $\min(v_p(b_i), v_p(b_j))$ from the LHS. We will only use the fact that if $p \nmid j-i$, then $\max(v_p(b_i), v_p(b_j)) \ge v_p(d)$ Similar to before, let $b_k$ denote the term with minimal $v_p$ and say $k \equiv r \pmod p$ then everything that is not $r \pmod p$ has $v_p$ at least $v_p(d)$, so $\frac{n(p-1)}{p} \le \frac{5n}{p^{v_p(d)}} \implies p^{v_p(d) - 1}(p-1) \le 5$. Taking $p = 5$, we have that $z \le 1$, similarly we get $y \le 1$ and $x \le 3$. So, $d \le (2^3)(3)(5) = 120$, as claimed. $\square$ Now, we will show that because $d$ is small, the entire sequence $\frac{a_i}{b_i}$ grows too fast to have only numbers $\le 5n$. We split into two cases: Case 1: $5 \nmid d$ Proof: In this case, we have $d \le 24$. So, $\frac{a_n}{b_n} > \frac{n}{24} > 84$. Consider all the terms that are $> 15$. There must be at least $n - 15(24) = n-360$ such terms. But for $\frac{a_i}{b_i} > 15$, we must have $a_i < \frac{5n}{15} = \frac{n}{3}$, so at most $\frac{n}{3}$ such terms can exist. So, $\frac{n}{3} > n-360 \iff n < 540$, which is not true. $\square$. Case 2: $5 | d$ Proof: In this case we only have $d \le 120$, however, we know that at least $\frac{4n}{5}$ of the $b_i$ are divisible by $5$. So, consider terms among these which are $> 2$. There must be at least $\frac{4n}{5} - 240$ such terms. The point is that since we have $5 | b_i$, there can be at most $\frac{5n}{10} = \frac{n}{2}$ terms. So, we must have $\frac{4n}{5} - 240 < \frac{n}{2} \iff n < 800$, a contradiction. $\square$ Since this is not possible, we must have had $\frac{c}{d} = 0$, so the entire sequence is just constant, as desired. $\blacksquare$

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My two hundredth post I reached my 199th post three months ago and wanted to use my 200th post for a more special problem.I believe I have found said problem. Both a very natural statement and a natural solution, a problem where trying to imagine what the final solution should look like may be in fact be detrimental. This problem took me approximately four times longer to write-up than to solve. I do not have any regrets and prefer both writing and reading solutions with motivational remarks. Our motto in this solution will be: "More bounds are better than stronger bounds". This is a problem in which quantity may outweigh quality and the main reason is that one can almost immediately conclude that the problem holds for all sufficiently large $n \in \mathbb{N}$ and therefore one almost reduces to a "finite case check". I have tried to include as much of the motivation for the steps. Even though this is not a good way of advertising a solution, the solution below is the weakest on the thread, in the sense that if $2018$ was replaced by $2017$, the proof doesn't immediately follow, we actually use the constant $2018$! Nevertheless, I am sorry about the length of this solution but hope that you will enjoy.

Assume for the sake of contradiction that we have a non-constant arithmetic progression satisfying the condition, let $\frac{a}{b} > 0$ for $gcd(a,b) = 1$ be the common difference, we have that $$\frac{a_i}{b_i} + \frac{a}{b} = \frac{a_{i+1}}{b_{i+1}}$$ $\color{blue} \boxed{\textbf{Approach 1: p-adic valuation}}$ As we are not able to get any algebraic constraints using the fact that the integers are at most $5n$, we are motivated consider the size constraint as a bound on the number of integers among the $2n$ integers that are divisible by some $k \in \mathbb{N}$, we shall therefore closely examine $\nu_p(\frac{a_i}{b_i})$. The first lemma is possibly one of the most natural constraints that one gets from the arithmetic progression, the above motivation immediately gives us $\textbf{Lemma 2}$ as we roughly have that if $k \mid a_{c_1}, \cdots, a_{c_m}, b_{d_1}, \cdots, b_{d_t}$ with $c_i \neq c_j$ and $d_i \neq d_j$ for $i \neq j$ in $\{1, \cdots, n\}$, then $k(m+t) \leq 5n$. $\textbf{Lemma 1:}$ If $\nu_p(b) > \nu_p(b_i)$ then $\nu_p(b_{i+1}) \geq \nu_p(b)$ for all $i \in \{1,2, \cdots, n-1\}$

$\textbf{Lemma 2:}$ $p^{\nu_p(b)} < 10$ for all primes $p$.

$\textbf{Corollary to Lemma 2:}$ $b = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta}$ for $\alpha \leq 3, \beta \leq 2, \gamma \leq 1, \delta \leq 1$. $\textbf{Lemma 3:}$ If for some prime $p \mid b$, there is some $i \in \{1,2, \cdots, n\}$ such that $$\nu_p(\frac{a_i}{b_i}) < \nu_p(\frac{a}{b})$$then $p=2$.

In order to preserve the chronological order in which I completed the steps, we have the following section where we try and lower bound $b$. $\color{red} \boxed{\textbf{Approach 2: Sub-sequence of Integral Difference in Rational Arithmetic Progression}}$ The motivation for this section is pretty clear and before explaining what the main idea will be, I shall present the following case, which will easily generalize to different $b \in \mathbb{N}$. $\textbf{Lemma 4:}$ It is not possible that $\frac{a}{b} \in \mathbb{N}$ in other words that $b = 1$.

We will slightly reformulate the sub-problem that is to be solved for the sake of simpler notation. $\textbf{Goal:}$ Let us find an upper bound on $M$ depending on $n \geq 2018$ where $M$ is the length of the longest possible sequence $\mu_1, \cdots, \mu_M$ such that $(\mu_u)_{i \geq 1}$ is an arithmetic progression with integral common difference and $\mu_i = \frac{x_i}{y_i}$ for some $x_i,y_i \in \{1, \cdots, 5n\}$ with all $2M$ integers distinct. $\textbf{Lemma 5:}$ $M \leq \sqrt{20n+1}$

We will now conclude that $b > 10$. $\textbf{Lemma 6:}$ $b > 10$

$\color{blue} \boxed{\textbf{Section 3: Algebraic Conclusions Based on p-adic Analysis}}$ It seems as though we have not gotten much out of $\color{red} \textbf{Approach 2}$, we shall see if $b > 10$ was much progress. Given the existence of $\textbf{Lemma 2}$ any sort of bound we can get should be useful after all. In order to get a better traction of the $\nu_p(\frac{a_i}{b_i})$ for each $i \in \{1, \cdots, n\}$ write $\frac{a_i}{b_i} = \frac{x_i}{y_i}$ with $x_i,y_i \in \mathbb{N}, gcd(x_i,y_i) = 1$. For what is to follow, we will assume for the sake of contradiction that $b \neq 12$. We wil analyze the case $b=12$ later. $\color{red} \boxed{\textbf{Section 4: Main Global Idea}}$ Before revealing the idea, which seems to be unique when compared to the previous posters as well as the solution on the shortlist packet, we will bound $b$ above because the current bound of $b \leq 2560$ by $\textbf{Lemma 1}$ is not sufficient as it is currently possible that $b>n$. $\textbf{Lemma 7: More Local Combinatorial Arguments}$ $b \leq \frac{2520}{6} = 420$

In order to have a more linear presentation of the solution, we will first bound $\varphi(b)$ and then apply our bounds to conclude the case when $b > 12$. $\textbf{Lemma 8:}$ If $m \mid 2^3 \cdot 3^2 \cdot 5 \cdot 7$ and $m > 12$, then $\varphi(m) \geq 6$

Now, applying $\textbf{Lemma 7}$ for $m=b$, we can notice that $\varphi(b) \geq 6$ as by the corollary of $\textbf{Lemma 2}$, we know that $b \mid 2^3 \cdot 3^2 \cdot 5 \cdot 7$. We will now split into two cases for the sake of clarity, even though the bounds that we get from one of the cases will be strictly stronger than the other, which will allow us to rule said case out. First of all, write $\frac{x_1}{y_1} = \frac{t}{kb}$ for some minimal $k \in \mathbb{N}$ and an appropriate $t \in \mathbb{N}$ and rewrite $\frac{a_i}{b_i} = \frac{t+ka(i-1)}{kb}$. By $\textbf{Lemma 3}$, $gcd(k,b) \mid 2$. $\textbf{Case 1)}$ $gcd(k,b) = 2$

$\textbf{Case 2:)}$ $gcd(k,b) = 1$ The same reasoning with the previous notation allows us to conclude that among each of the $\lfloor \frac{n}{b} \rfloor$ $b$-tuplets, $\{1,2, \cdots, n\}$ into subsets $\{1, \cdots, b\}, \{b+1, \cdots, 2b\}, \cdots, \{(\lfloor \frac{n}{b} \rfloor -1)b+1, \cdots, \lfloor \frac{n}{b} \rfloor \cdot b\}$ contain at least $\phi(b)$ integers such that $b \mid b_i$ this gives us the bound $b\phi(b) \cdot \lfloor \frac{n}{b} \rfloor \leq 5n$, as this bound is weaker than that we got in $\textbf{Case 1}$, we will analyze this inequality instead. $$b\phi(b) \cdot (\frac{n}{b}-1) \leq b\phi(b) \cdot \lfloor \frac{n}{b} \rfloor \leq 5n$$after rearranging this gives us that $$n \leq \frac{b\phi(b)}{\phi(b)-5} \leq 6b$$because $\phi(b) \geq 6$ by $\textbf{Lemma 8}$. Then as $b \mid 2520$ and $b \leq 420$ by $\textbf{Lemma 2}$ and $\textbf{Lemma 7}$, we get that $b \in \{360,420\}$ because $2018 \leq n \leq 6b$, in these cases it is easy to verify that $n \leq b\phi(b)(\phi(b)-5) < 2018$, a contradiction. As mentioned before, this concludes $\textbf{Case 1}$ as well. $\blacksquare$ We have dealt with $b \neq 12$! $\color{red} \boxed{\textbf{Specificity Allows for Improvements}}$ Even though none of the previous methods were able to eliminate the case $b=12$ due to the fact that $\phi(12) = 4$, we will not struggle in dealing with the case $b=12$ because it is a very specific case and we should intuitively be able to exploit the constant itself to improve various bounds. In the way that it has previously been done in $\color{red} \textbf{Approach 2}$, we will again look at sub-sequences of integral difference but this time rather than focusing on the sub-sequence $\frac{a_1}{b_1}, \frac{a_{12+1}}{b_{12+1}}, \frac{a_{2 \cdot 12+1}}{b_{2 \cdot 12+1}}, \cdots$, we will look at each of the $12$ sub-sequences created by starting at $\frac{a_i}{b_i}$ for $i \in \{1, \cdots, 12\}$. $\textbf{Lemma 9:}$ $b \neq 12$

Ultimately, $\textbf{Approach 2}, \textbf{Section 3}$ and $\textbf{Lemma 9}$ exhaust all possible cases for $b > 0$ and due to the contradictions in each of the cases, we must have that the sequence is constant. $\blacksquare$ I will now again discuss the motivation behind the solution, these remarks will be a lot more general than those I have included before the individual lemmas.

Let the common difference of the arithmetic progression be $\frac{c}{d}$ (FTSOC assume that $c\neq 0$) where $\gcd(c,d)=1$. WLOG assume that $\frac{c}{d}>0$. Claim 1. We must have $d>\frac{n}{20}>100$. Proof: We have that $\frac{a_{i+1}}{b_{i+1}}=\frac{a_{i}}{b_{i}}+\frac{c}{d}\geq \frac{a_{i}}{b_{i}}+\frac{1}{d}$ for all $i$, so $\frac{a_{i}}{b_{i}}\geq \frac{i-1}{d}+\frac{a_{1}}{b_{1}}\geq \frac{i-1}{d}+\frac{1}{5n}$ by induction. Now consider the set (here $s=\lfloor\frac{n+1}{2}\rfloor$): \[\left\{\frac{a_{s}}{b_{s}},\frac{a_{s+1}}{b_{s+1}},\dots ,\frac{a_{n}}{b_{n}}\right\}\]and let $b_{j}=\max_{s\leq i\leq n} b_{i}$. As there are $n-s+1$ different positive integers among $b_{s},b_{s+1},\dots ,b_{n}$, we get $b_{j}\geq n-s+1$. Now we have the following inequality chain: \[\frac{s-1}{d}+\frac{1}{5n}\leq \frac{j-1}{d}+\frac{1}{5n}\leq\frac{a_{j}}{b_{j}}\leq \frac{5n}{n-s+1}\]We examine two cases depending on the parity of $n$: $n$ is even. Then $n=2s$ and the inequality above transforms into: \[\frac{s-1}{d}+\frac{1}{10s}\leq \frac{10s}{s+1}\]\[\Longrightarrow 10s(s^2-1)+(s+1)d\leq 100s^2d\]\[\Longrightarrow d\geq \frac{10s^3-10s}{100s^2-s-1}\geq \frac{s}{10}=\frac{n}{20}\geq \frac{2018}{20}>100\] $n$ is odd. Then $n=2s-1$ and the inequality above transforms into: \[\frac{s-1}{d}+\frac{1}{10s-5}\leq \frac{10s-5}{s}\]\[\Longrightarrow s(s-1)(10s-5)+ds\leq d(10s-5)^2\]\[\Longrightarrow d\geq \frac{s(s-1)(10s-5)}{(10s-5)^2-s}\geq \frac{s-\frac{1}{2}}{10}=\frac{n}{20}>100\]We got that $d\geq \frac{n}{20}$ in both cases, so we're done. Now because $d>1$ we can write $d=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{k}^{\alpha_{k}}$. Claim 2. We must have $p_{i}\leq 5$ for all $i$. Proof: Assume that $p\mid d$ for some prime $p\geq 7$. Then at least $n-\lceil\frac{n}{p}\rceil$ of the $n$ simplified fractions have a denominator divisible by $p$. Indeed, in the set $\left\{\frac{a_{i}}{b_{i}},\frac{a_{i+1}}{b_{i+1}},\dots ,\frac{a_{i+p-1}}{b_{i+p-1}}\right\}$ at most one fraction has a denominator not divisible by $p$. If we assume otherwise, that means that there exist $i\leq j_{1}<j_{2}\leq i+p-1$ such that $p$ doesn't divide the denominator in the simplified fractions $\frac{a_{j_{1}}}{b_{j_{1}}}$ and $\frac{a_{j_{2}}}{b_{j_{2}}}$, so: \[0\leq \min\left\{v_{p}\left(\frac{a_{j_{1}}}{b_{j_{1}}}\right), v_{p}\left(\frac{a_{j_{2}}}{b_{j_{2}}}\right)\right\}\leq v_{p}\left(\frac{a_{j_{1}}}{b_{j_{1}}}-\frac{a_{j_{2}}}{b_{j_{2}}}\right)=v_{p}\left(\frac{c(j_{2}-j_{1})}{d}\right)=\]\[=v_{p}(c)+v_{p}(j_{2}-j_{1})-v_{p}(d)=-v_{p}(d)<0\]as $p\mid d$, $\gcd(c,d)=1\Longrightarrow p\nmid c$ and $0<j_{2}-j_{1}<p\Longrightarrow v_{p}(j_{2}-j_{1})=0$. Therefore there are at most $\lceil\frac{n}{p}\rceil$ of those $n$ simplified fractions which don't have a denominator divisible by $p$. Now, the fractions which have a $p$ in their simplified denominator must have $p\mid b_{i}$. However, there are only $\lfloor\frac{5n}{p}\rfloor$ such integers between $1$ and $5n$ inclusive, therefore: \[n-\left\lceil\frac{n}{p}\right\rceil\leq \left\lfloor\frac{5n}{p}\right\rfloor\]Using the weak inequalities $n-\left\lceil\frac{n}{p}\right\rceil\geq n-\frac{n}{p}-\frac{p-1}{p}$ and $\lfloor\frac{5n}{p}\rfloor\leq \frac{5n}{p}$ we get: \[n-\frac{n}{p}-\frac{p-1}{p}\leq n-\left\lceil\frac{n}{p}\right\rceil\leq \left\lfloor\frac{5n}{p}\right\rfloor\leq \frac{5n}{p}\]\[\Longrightarrow 0\leq (6-p)n+p-1\]However, $p\geq 7$, so $6-p<0$ and thus we have: \[(6-p)n+p-1\leq (6-p)2018+p-1=12107-2017p\leq 12107-7\cdot 2017=-2012<0\]which is a contradiction with out assumption, so the proof is complete. Let $d=2^{\alpha_{1}}3^{\alpha_{2}}5^{\alpha_{3}}$. Note that from the first claim we know that $d>100$, so we must have $\alpha_{1}\geq 3$, $\alpha_{2}\geq 2$ or $\alpha_{3}\geq 2$ as otherwise $d\leq 2^2.3.5=60<100$. We consider the following three cases: Case 1. $\alpha_{1}\geq 4$. Then at least $n-\lceil\frac{n}{2}\rceil$ of the $n$ simplified fractions have a denominator divisible by $8$. Indeed, in the set $\left\{\frac{a_{i}}{b_{i}},\frac{a_{i+1}}{b_{i+1}}\right\}$ at most one fraction has a denominator not divisible by $16$. If we assume otherwise, then $16$ doesn't divide the denominators in the simplified fractions $\frac{a_{i}}{b_{i}}$ and $\frac{a_{i+1}}{b_{i+1}}$, so: \[-3\leq \min\left\{v_{2}\left(\frac{a_{i}}{b_{i}}\right), v_{2}\left(\frac{a_{i+1}}{b_{i+1}}\right)\right\}\leq v_{2}\left(\frac{a_{i}}{b_{i}}-\frac{a_{i+1}}{b_{i+1}}\right)=v_{2}\left(\frac{c(i+1-i)}{d}\right)=\]\[=v_{2}(c)-v_{2}(d)\leq -4\]as $16\mid d$, $\gcd(c,d)=1\Longrightarrow 2\nmid c$. Therefore there are at most $\lceil\frac{n}{2}\rceil$ of those $n$ simplified fractions which don't have a denominator divisible by $16$. Now, the fractions which have a $16$ in their simplified denominator must have $16\mid b_{i}$. However, there are only $\lfloor\frac{5n}{16}\rfloor$ such integers between $1$ and $5n$ inclusive, therefore: \[n-\left\lceil\frac{n}{2}\right\rceil\leq \left\lfloor\frac{5n}{16}\right\rfloor\Longrightarrow n-\frac{n}{2}-\frac{1}{2}\leq n-\left\lceil\frac{n}{2}\right\rceil\leq \left\lfloor\frac{5n}{16}\right\rfloor\leq \frac{5n}{16}\Longrightarrow 3n\leq 8\]However, $n\geq 2018$, so this is obviously wrong. Case 2. $\alpha_{2}\geq 2$. Then at least $n-\lceil\frac{n}{3}\rceil$ of the $n$ simplified fractions have a denominator divisible by $9$. Indeed, in the set $\left\{\frac{a_{i}}{b_{i}},\frac{a_{i+1}}{b_{i+1}},\frac{a_{i+2}}{b_{i+2}}\right\}$ at most one fraction has a denominator not divisible by $9$. If we assume otherwise, that means that there exist $i\leq j_{1}<j_{2}\leq i+2$ such that $9$ doesn't divide the denominator in the simplified fractions $\frac{a_{j_{1}}}{b_{j_{1}}}$ and $\frac{a_{j_{2}}}{b_{j_{2}}}$, so: \[-1\leq \min\left\{v_{3}\left(\frac{a_{j_{1}}}{b_{j_{1}}}\right), v_{3}\left(\frac{a_{j_{2}}}{b_{j_{2}}}\right)\right\}\leq v_{3}\left(\frac{a_{j_{1}}}{b_{j_{1}}}-\frac{a_{j_{2}}}{b_{j_{2}}}\right)=v_{3}\left(\frac{c(j_{2}-j_{1})}{d}\right)=\]\[=v_{3}(c)+v_{3}(j_{2}-j_{1})-v_{3}(d)\leq -2\]as $9\mid d$, $\gcd(c,d)=1\Longrightarrow 3\nmid c$ and $0<j_{2}-j_{1}<3\Longrightarrow v_{3}(j_{2}-j_{1})=0$. Therefore there are at most $\lceil\frac{n}{3}\rceil$ of those $n$ simplified fractions which don't have a denominator divisible by $9$. Now, the fractions which have a $9$ in their simplified denominator must have $9\mid b_{i}$. However, there are only $\lfloor\frac{5n}{9}\rfloor$ such integers between $1$ and $5n$ inclusive, therefore: \[n-\left\lceil\frac{n}{3}\right\rceil\leq \left\lfloor\frac{5n}{9}\right\rfloor\Longrightarrow n-\frac{n}{3}-\frac{2}{3}\leq n-\left\lceil\frac{n}{3}\right\rceil\leq \left\lfloor\frac{5n}{9}\right\rfloor\leq \frac{5n}{9}\Longrightarrow n\leq 6\]However, $n\geq 2018$, so this is obviously wrong. Case 3. $\alpha_{3}\geq 2$. Then at least $n-\lceil\frac{n}{5}\rceil$ of the $n$ simplified fractions have a denominator divisible by $25$. Indeed, in the set $\left\{\frac{a_{i}}{b_{i}},\frac{a_{i+1}}{b_{i+1}},\dots ,\frac{a_{i+4}}{b_{i+4}}\right\}$ at most one fraction has a denominator not divisible by $25$. If we assume otherwise, that means that there exist $i\leq j_{1}<j_{2}\leq i+4$ such that $25$ doesn't divide the denominator in the simplified fractions $\frac{a_{j_{1}}}{b_{j_{1}}}$ and $\frac{a_{j_{2}}}{b_{j_{2}}}$, so: \[-1\leq \min\left\{v_{5}\left(\frac{a_{j_{1}}}{b_{j_{1}}}\right), v_{5}\left(\frac{a_{j_{2}}}{b_{j_{2}}}\right)\right\}\leq v_{5}\left(\frac{a_{j_{1}}}{b_{j_{1}}}-\frac{a_{j_{2}}}{b_{j_{2}}}\right)=v_{5}\left(\frac{c(j_{2}-j_{1})}{d}\right)=\]\[=v_{5}(c)+v_{5}(j_{2}-j_{1})-v_{5}(d)\leq -2\]as $25\mid d$, $\gcd(c,d)=1\Longrightarrow 5\nmid c$ and $0<j_{2}-j_{1}<5\Longrightarrow v_{5}(j_{2}-j_{1})=0$. Therefore there are at most $\lceil\frac{n}{5}\rceil$ of those $n$ simplified fractions which don't have a denominator divisible by $25$. Now, the fractions which have a $25$ in their simplified denominator must have $25\mid b_{i}$. However, there are only $\lfloor\frac{n}{5}\rfloor$ such integers between $1$ and $5n$ inclusive, therefore: \[n-\left\lceil\frac{n}{5}\right\rceil\leq \left\lfloor\frac{n}{5}\right\rfloor\Longrightarrow n-\frac{n}{5}-\frac{4}{5}\leq n-\left\lceil\frac{n}{5}\right\rceil\leq \left\lfloor\frac{n}{5}\right\rfloor\leq \frac{n}{5}\Longrightarrow 3n\leq 4\]However, $n\geq 2018$, so this is obviously wrong. As a result, we get that $\alpha_{1}\leq 3$, $\alpha_{2}\leq 1$, $\alpha_{3}\leq 1$. If any of these inequalities isn't an equality, then $d<100$, so we must have $d=2^3.3.5=120$. However, now we do the same trick for $15$: Case 4. $\alpha_{2}=1$ and $\alpha_{3}=1$. Then at least $n-\lceil\frac{n}{3}\rceil$ of the $n$ simplified fractions have a denominator divisible by $15$. Indeed, in the set $\left\{\frac{a_{i}}{b_{i}},\frac{a_{i+1}}{b_{i+1}},\frac{a_{i+2}}{b_{i+2}}\right\}$ at most one fraction has a denominator not divisible by $15$. If we assume otherwise, that means that there exist $i\leq j_{1}<j_{2}\leq i+2$ such that $15$ doesn't divide the denominator in the simplified fractions $\frac{a_{j_{1}}}{b_{j_{1}}}$ and $\frac{a_{j_{2}}}{b_{j_{2}}}$, so: \[0\leq \min\left\{v_{15}\left(\frac{a_{j_{1}}}{b_{j_{1}}}\right), v_{15}\left(\frac{a_{j_{2}}}{b_{j_{2}}}\right)\right\}\leq v_{15}\left(\frac{a_{j_{1}}}{b_{j_{1}}}-\frac{a_{j_{2}}}{b_{j_{2}}}\right)=v_{15}\left(\frac{c(j_{2}-j_{1})}{d}\right)=\]\[=v_{15}(c(j_{2}-j_{1}))-v_{15}(120)=-1\]as $15\mid d$, $\Longrightarrow\gcd(c,15)=1$ and $0<j_{2}-j_{1}<3\Longrightarrow \gcd(15,j_{2}-j_{1})=1$. Therefore there are at most $\lceil\frac{n}{3}\rceil$ of those $n$ simplified fractions which don't have a denominator divisible by $15$. Now, the fractions which have a $15$ in their simplified denominator must have $15\mid b_{i}$. However, there are only $\lfloor\frac{n}{3}\rfloor$ such integers between $1$ and $5n$ inclusive, therefore: \[n-\left\lceil\frac{n}{3}\right\rceil\leq \left\lfloor\frac{n}{3}\right\rfloor\Longrightarrow n-\frac{n}{3}-\frac{2}{3}\leq n-\left\lceil\frac{n}{3}\right\rceil\leq \left\lfloor\frac{n}{3}\right\rfloor\leq \frac{n}{3}\Longrightarrow n\leq 2\]However, $n\geq 2018$, so this is obviously wrong. Therefore we reached a contradiction with the result from the first claim and so $c=0$.

We start by considering the $p$-adic structure of arithmetic progressions. Claim 1: Let $c_1,\ldots,c_m$ be an arithmetic progression with common difference $d$. If $p$ is a prime and $k=\min(\nu_p(d),\nu_p(c_1))$, then every term of the sequence has a $p$-adic valuation of $k$ or exactly one of every $p$ consecutive terms has a $p$-adic valuation other than $k$. Proof: If $\nu_p(d)>\nu_p(c_1)$, then all terms of the sequence have a $p$-adic valuation of $\nu_p(c_1)$. If $\ell=\nu_p(d) \le \nu_p(c_1)$, then $p^{-\ell} d$ exists and is nonzero modulo $p$. Thus, $0,p^{-\ell} d,\ldots,(p-1)p^{-\ell} d$ forms a complete residue class modulo $p$, so exactly one of \[p^{-\ell} c_i,p^{-\ell} c_i+p^{-\ell} d,\ldots,p^{-\ell} c_i+(p-1)p^{-\ell} d\]is divisible by $p$ for $i=1,\ldots,m$. It follows that exactly one of $c_i,c_i+d,\ldots,c_i+(p-1)d$ has a $p$-adic valuation not equal to $p^\ell$. $\square$ For any prime $p$, let $k_p$ be the value of $k$ for the sequence $\tfrac{a_1}{b_1},\ldots,\tfrac{a_n}{b_n}$ given Claim 1. Claim 2: $k_2 \ge -3$, $k_3 \ge -1$, $k_5 \ge -1$, and $k_p \ge 0$ for primes $p \ge 7$. Proof: For a prime $p$ with $k_p \le 0$, at least $\lfloor \tfrac{(p-1)n}{p} \rfloor$ terms of $b_1,\ldots,b_n$ are divisible by $p^{-k_p}$ by Claim 1. Since there are at most $\tfrac{5n}{p^{-k_p}}$ positive multiples of $p^{-k_p}$ not exceeding $5n$, we have $\tfrac{5n}{p^{-k_p}} \ge \lfloor \tfrac{(p-1)n}{p} \rfloor \ge \tfrac{(p-1)n}{p}-1$. This is equivalent to \[\left(\frac{p-1}{p}-\frac{5}{p^{-k_p}}\right)n \le 1.\]Notice that for $p=2$, $-k_p \ge 4$ gives $\tfrac{3}{16}n \le 1$ for $p=3$, $-k_p \ge 2$ gives $\tfrac{1}{9}n \le 1$ for $p=5$, $-k_p \ge 2$ gives $\tfrac{3}{5}n \le 1$ for $p \ge 7$, $-k_p \ge 1$ gives $\tfrac{p-6}{p}n \le 1$, a contradiction. $\square$ Before considering the common difference of $\tfrac{a_1}{b_1},\ldots,\tfrac{a_n}{b_n}$, we deal with an edge case. Claim 3: $(k_2,k_3,k_5) \ne (-3,-1,-1)$. Proof: This proof imitates that of Claim 2 with $120$ instead of a prime power. Assume FTSOC that $(k_2,k_3,k_5)=(-3,-1,-1)$. By Claim 1, at least $\varphi(30)=8$ of every $30$ consecutive terms have a $2$-adic valuation of $-3$, a $3$-adic valuation of $-1$, and a $5$-adic valuation of $-1$. This means there are at least $8\lfloor \tfrac{n}{30} \rfloor \ge \tfrac{4}{15}n-8$ indices $i$ such that $b_i$ is divisible by $120$. We also have an upper bound of $\tfrac{5}{120}n$ positive multiples of $120$ not exceeding $5n$, so we have $\tfrac{4}{15}n-8 \le \tfrac{5}{120}n \implies \tfrac{9}{40}n \le 8$, a contradiction. $\square$ Let $\tfrac{c}{d}$ be the common difference of $\tfrac{a_1}{b_1},\ldots,\tfrac{a_n}{b_n}$ when written in lowest terms. Assume WLOG that $c \ge 0$, and assume FTSOC that $c \ne 0$. Claim 2 gives $d \mid 120$ and Claim 3 gives $d \ne 120$, so we have $d \le 60$. Notice that at least one of $b_{\lceil n/2 \rceil},\ldots,b_n$ is at least $\tfrac{n}{2}$ and $\tfrac{a_i}{b_i} \ge (i-1)d \ge \tfrac{\frac{n}{2}-1}{120}$ for $i=\lfloor \tfrac{n}{2} \rfloor,\ldots,n$. Therefore, at least one of $a_{\lfloor n/2 \rfloor},\ldots,a_n$ is at least \[\frac{n}{2} \cdot \frac{\frac{n}{2}-1}{60} \ge \frac{n}{2} \cdot \frac{\frac{n}{3}}{60}=\frac{n^2}{360}>5n,\]a contradiction. Thus, we must have $c=0$, as desired. $\square$

Suppose otherwise. WLOG assume the arithmetic sequence is increasing. Let $r_i = \frac{a_i}{b_i}$ and the common difference of the arithmetic progression be $d$. Claim: $120d$ is an integer. Proof: It suffices to prove that $\nu_2(d) \ge -3, \nu_3(d) \ge -1, \nu_5(d) \ge -1$, and $\nu_p(d) \ge 0$ for all primes $p > 5$. Suppose for some $p$ this wasn't true (as in $\nu_p(d) < -3$ if $p = 2$, etc). Case 1: $\nu_p(r_1) < \nu_p(d)$. Then\[\nu_p(r_1) = \nu_p(r_2) = \cdots = \nu_p(r_n) < \nu_p(d)\]Thus, $b_i$ must be divisible by $p^{-\nu_p(d) + 1} $ for each $i$. Since the $b_i$ are pairwise distinct, we must have $n \cdot p^{-\nu_p(d) + 1} $ be at most $5n$ (or else one of the $b_i$ would exceed $5n$), so $p^{-\nu_p(d) + 1} \le 5$. Now if $\nu_p(d) < 0$, we must have that $p^2 \le 5$, so $p = 2$. If $p = 2$, then $-\nu_p(d) + 1$ can't exceed $2$, so $\nu_p(d) \ge -1 \ge -3$, as desired. Case 2: $\nu_p(r_1) \ge \nu_p(d)$. We see that all terms of the arithmetic sequence have a $\nu_p$ at least equal to $\nu_p(d)$. Consider any $p$ consecutive terms of the arithmetic sequence, say $r_i, r_i + d,\ldots, r_i + d(p-1)$. Notice that if any two had a $\nu_p$ greater than $\nu_p(d)$, then by subtracting them, we get a contradiction as their difference is $d$ times a non multiple of $p$. Hence within any $p$ consecutive terms of the arithmetic sequence, at most one of them has a $\nu_p$ greater than $\nu_p(d)$. This implies that at least $n - \left \lceil \frac np \right \rceil $ numbers in the arithmetic sequence have a $\nu_p$ equal to that of $\nu_p(d)$. This implies that at least $n - \left \lceil \frac np \right \rceil $ of $b_1, b_2, \ldots, b_n$ are multiples of $p^{- \nu_p(d)}$, so at least one of them is at least $p^{-\nu_p(d)} \cdot \left( n - \left \lceil \frac np \right \rceil \right) > p^{-\nu_p(d)} \cdot \left( n - \frac np - 1 \right) $, implying that\[ p^{ - \nu_p(d) } \cdot \left(n - \frac np - 1 \right) < 5n \]The LHS is equal to $n (p^{ - \nu_p(d)} - p^{ - \nu_p(d) - 1} ) - p^{ - \nu_p(d)} $, so the inequality rearranges to \[ n \left( p^{ - \nu_p(d)} - p^{- \nu_p(d) - 1} - 5 \right) < p^{- \nu_p(d)} \]Now let $x = - \nu_p(d)$. Assume $x > 0$ as otherwise the claim would obviously be true. Now we claim that the LHS must be negative. If not, we would have that $2018 \le n < \frac{p^x}{p^x - p^{x-1} - 5}$. But the reciprocal of this expression is $1 - \frac 1p - \frac{5}{p^x}$ and must be less than $\frac{1}{2018}$. If $p = 2$, then since $x \le 4$, $1 - \frac 1p - \frac{5}{p^x} \ge 1 - \frac 12 - \frac{5}{16} > \frac{1}{2018}$. Now if $p \in \{3,5\} $, then since $x \le 2$, we have $1 - \frac 1p - \frac{5}{p^x} \ge 1 - \frac 13 - \frac{5}{3^2} > \frac{1}{2018}$. Otherwise, if $p > 5$, then $1 - \frac 1p - \frac{5}{p^x} \ge 1 - \frac 17 - \frac{5}{7} > \frac{1}{2018}$. Hence the LHS must be negative. This implies that $p^x - p^{x-1} - 5 < 0$, so $p^{x-1} (p - 1) < 5$. For $p > 5$, we have $p^{x-1} < 1$, so $x < 1$, contradiction. If $p = 5$, then $5^{x-1} < \frac 54$, so $x < 2$, meaning that $\nu_5(d) \ge -1$. If $p = 3$, then $3^{x-1} < \frac 52$, so $x < 2$, so $\nu_3(d) \ge -1$. If $p = 2$, then $2^{x-1} < 5$, so $x < 4$, meaning $\nu_2(d) \ge -3$, as desired. $\square$ Claim: At least one of $\nu_3(d)$ and $\nu_5(d)$ is nonnegative. Proof: Suppose otherwise and $\nu_3(d) = \nu_5(d) = -1$. Recall that if $\nu_p(r_1) < \nu_p(d)$, then $p^{ - \nu_p(d) + 1} \le 5$, which is impossible for $p = 3,5$ in this case. Hence $\nu_p(r_1) \ge \nu_p(d)$ for $p\in \{3,5\}$. We copy this again (for any prime $p$ with $\nu_p(x_1)$ ): Consider any $p$ consecutive terms of the arithmetic sequence, say $r_i, r_i + d,\ldots, r_i + d(p-1)$. Notice that if any two had a $\nu_p$ greater than $\nu_p(d)$, then by subtracting them, we get a contradiction as their difference is $d$ times a non multiple of $p$. Hence within any $p$ consecutive terms of the arithmetic sequence, at most one of them has a $\nu_p$ greater than $\nu_p(d)$. This implies that at least $n - \left \lceil \frac np \right \rceil \ge n - \frac np - 1$ numbers in the arithmetic sequence have a $\nu_p$ equal to that of $\nu_p(d)$. Hence at most $\frac n3 + 1$ numbers in the sequence have a $\nu_3$ over $-1$ and at most $\frac n5 + 1$ numbers in the sequence have a $\nu_5$ over $-1$, implying that at least $1 - \left( \frac n3 + 1 + \frac n5 + 1 \right) = \frac{8}{15} \cdot n - 1$ numbers in the sequence have a $\nu_3$ and $\nu_5$ equal to $-1$. Hence these numbers must have a denominator divisible by $15$, so at least $\frac{8}{15} n - 1$ of the $b_i$ are multiples of $15$, which means some $b_k$ is at least $15 \cdot \left( \frac{8}{15} \cdot n - 1 \right) = 8n - 15 > 5n$ (for $n \ge 2018$), absurd. $\square$ These two claims imply that when $d$ is written as a simplified fraction of two integers (since $d$ is nonzero), the denominator is a divisor of $120$, but not a multiple of $15$, so it is at most $40$. Hence $d \ge \frac{1}{40}$. Now let $m = \left \lceil \frac n2 \right \rceil $. Notice that at least one number in $b_m, b_{m+1} , \ldots, b_n$ is at least $n - m + 1$. Let $b_k \ge n - m + 1$ for some $k\in [m,n]$. We have $r_k \le \frac{5n}{n - m + 1}$, so the common difference of the sequence is\[d = \frac{r_k - r_1}{k - 1 } < \frac{r_k}{m - 1} < \frac{5n}{(n - m + 1) (m-1)} \]Now if $n$ is even, let $n = 2t$, so $m = t$ and $t \ge 1009$. The value of $d$ is at most\[ \frac{10t}{(t + 1)(t - 1)} < \frac{10(t+1)}{(t+1)(t-1)} = \frac{10}{t-1} < \frac{10}{1008} , \]which is much less than $\frac{1}{40}$, absurd. If $n$ is odd, let $n = 2t - 1$, where $t \ge 1010$ and $m = t$. The value of $d$ is at most\[ \frac{10t - 5}{t(t-1)} < \frac{10t}{t(t-1)} = \frac{10}{t-1} < \frac{10}{1009},\]which is much less than $\frac{1}{40}$, absurd. Therefore, the arithmetic sequence must be constant.

$5n$ can be improved to $\mathcal O\left(\sqrt{\frac{n^3}{\ln\ln n}}\right).$

Here goes nothing. Let $\frac{c}d$ be the common difference with $\gcd(c,d)=1$ and assume $c>0$, and let $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_n\}$. So we have \[\frac{a_i}{b_i}+\frac{c}d(j-i) \iff c(j-i)b_ib_j=d(a_jb_i-a_ib_j) \implies d \mid (j-i)b_ib_j \implies d \mid b_ib_{i+1} \left(\clubsuit \right)\]Claim: We have $d \mid 2^6 \cdot 3^4 \cdot 5^2 \cdot 7^2$. Proof: Say $p \mid d$ and $\nu_p(d)=e$ and so \[p^e \mid b_ib_{i+1} \implies p^{\left \lceil \frac{e}2 \right \rceil} \mid \text{lcm}(b_i,b_{i+1}) \]And so there are minimum $\frac{n-1}2$ multiples of $p^{\left \lceil \frac{e}2 \right \rceil}$ among $B$ and hence \[p^{\left \lceil \frac{e}2 \right \rceil} \left(\frac{n-1}2 \right) \leq 5n\]And now by some trivial size observations we are done. $\blacksquare$. Claim: We have $d \mid 2^3 \cdot 3^2 \cdot 5 \cdot 7$. Proof: Say $7^2 \mid d$. Now we have \[\nu_7(b_i)+\nu_7(b_{i+1})=\nu_7(d)+\nu_7(a_{i+1}b_i-a_ib_{i+1})\]And hence if there exists $i$ such that $\nu_7(b_i)=\nu_7(b_{i+1})=1$, then $\nu_7(\text{LHS})=2$ but $\nu_7(\text{RHS}) \geq 3$. And so again we get that there are atleast $\frac{n-1}2$ multiples of $7^2$ among elements in $B$. Again this means we get \[49 \left(\frac{n-1}2 \right) \leq 5n\]and contradiction. We can use similar arguments to show that $2^4 \nmid d$, $3^3 \nmid d$ and $5^2 \nmid d$. $\blacksquare$ Claim: We have $7 \nmid d$. Proof: Say $7 \mid d$. Since $7 \mid d \mid (j-i)b_ib_j$, and so if $7 \nmid b_k$ for some $k$ then \[\ell \equiv k \pmod 7 \text{ for all } 7 \nmid b_{\ell}\]And hence there are atmost $\left \lceil \frac{n}7 \right \rceil$ non multiples of $7$ among $B$ and hence there are atleast $\left \lfloor\frac{6n}7 \right \rfloor$ multiples of $7$ among $B$. And hence \[7 \left(\frac{6n}7-1 \right) \leq 5n\]and again contradiction. $\blacksquare$ Claim: We have $9 \nmid d$. Proof: Assume not, then we have $9 \mid d \mid (j-i)b_ib_j$. $\bullet$ Finding $k$ such that $\nu_3(k)=0$. Say $3 \nmid b_k$. Then $\ell \equiv k \pmod 9$ for all $3 \nmid b_{\ell}$. And hence there are atmost $\left \lceil \frac{n}9 \right \rceil$ non-multiples of $3$ among $B$. $\bullet$ Finding $k$ such that $\nu_3(k)=1$. Say for some $i$ and $j$ we have $\nu_3(b_i)=\nu_3(b_j)=1$ and $3 \nmid j-i$. Then now referring to the main equation in $\clubsuit$, we get that $\nu_3(\text{LHS})=2$ but $\nu_3(\text{RHS}) \geq 3$, contradiction. And so we get that there are atmost $\left \lceil \frac{n}3 \right \rceil$ elements among $B$ with $3$-adic evaluation as $1$. Hence we get that are atleast \[n-\left \lceil \frac{n}3 \right \rceil-\left \lceil \frac{n}9 \right \rceil \geq \frac{5n}9-1 \text{ multiples of $9$ among $B$}\]But see that there maximum $\frac{5n}9$ multiples of $9$ among $\{1,2,\dots,5n\}$. And hence there is maximum $1$ multiple of $9$ among $A$. Say we have $\nu_3(b_j) \geq \nu_3(b_i)+2 \geq 4$ for some $i$, $j$ such that $\nu_3(a_j) \neq 2$ And again referring to the main equation in $\clubsuit$, we get that \[\nu_3(\text{LHS}) \geq 2\nu_3(b_i)+2 \text{ but } \nu_3(\text{RHS}) \leq \nu_3(b_i)+3\]which is a contradiction. Hence is there are $N$ multiplies of $9$ in $B$, then for $N-1$ of those numbers we have that their $3$-adic evaluation can be one of two consecutive fixed numbers (which must be $2$ and $3$ or else size issues), and hence there are maximum $2$ multiples of $81$ among $\{1,2\dots,n\}$, which is obviously false. $\blacksquare$ And hence we get that \[d \mid 2^3 \cdot 3 \cdot 5=120\]Now we take two cases. $\bullet$ If $\frac{c}d \geq \frac1{60}$. See that we have \[\frac{a_i}{b_i} \geq \frac{i-1}{60} \implies \frac{a_n}{b_n} \geq \frac{n-1}{60} \geq 32\]See that there is minimum $n-600$ numbers $t$ such that $\frac{a)t}{b_t}>10$, and say $b_T$ is the maximum of those numbers and we get that \[5n \geq a_T>10b_T \geq 10(n-600) \implies n<1200\]which is a contradiction. $\bullet$ If $\frac{c}d < \frac1{60}$. See that we must have $d=120$ and so $\frac{c}d=\frac1{120}$; and \[\frac{a_i}{b_i} \geq \frac{i-1}{120} \implies \frac{a_n}{b_n} \geq \frac{n-1}{120} \geq 16\]Now using similar arguments before we get that if we look at the set of indices $t$, such that $\frac{a_t}{b_t}>2$, then the set of $i$ for which $5 \nmid b_i$ must be in the same residue class modulo $5$ and hence there atleast \[\left \lfloor \frac45 (n-240) \right \rfloor\]indices in set of such of $t$ such that $5 \mid b_t$. Let $b_T$ be the maximum of them all and we get \[5n \geq a_T \geq 2b_T \geq 2(4(n-240)-5) \implies n \leq 644\]which is a contradiction, and we are done. Well, that was a ride.

Absolutely beautiful question that explores multiple ways of attacking a problem. Lemma: Given a arithmetic sequence $q_1, q_2, \cdots q_n \in \mathbb Q$, if for some prime $p$ we have that $\text{min} \{\nu_p(q_1), \nu_p(q_2), \cdots, \nu_p(q_n) \} =\alpha$ then either there's $\left\lceil \frac{n}{p^{\ell}} \right\rceil$ terms of the sequence satisfy $\nu_p(q_i) \ge \alpha+\ell$. Or all terms are below this $\nu_p$. Proof: Perform a proper scaling so that $q_i \in \mathbb Z$ for all $1 \le i \le n$ and $\gcd(q_1, q_2, \cdots q_n)=1$, now let $d$ the difference of this sequence then if $p^{\ell} \mid q_i,q_j$ we get $p^{\ell} \mid q_i-q_j=d(i-j)$, now if $p \mid d$ then the minimun $\nu_p$ is no longer $0$ therefore we must have $p^{\ell} \mid i-j$ so such numbers are closed in a residue system $\pmod{p^{\ell}}$ which is enough to prove the Lemma. Claim: If $p^k \ge 7$ for $p^k$ being an odd prime power then $\nu_p \left(\frac{a_i}{b_i} \right) \ge 1-k$ for all $1 \le i \le n$ Proof: Suppose FTSOC that $\text{min} \left\{\nu_p \left(\frac{a_1}{b_1}\right), \nu_p \left(\frac{a_2}{b_2} \right), \cdots, \nu_p \left(\frac{a_n}{b_n} \right) \right\} \le -k$, from the lemma we know that at most $\left\lceil \frac{n}{p^k} \right\rceil$ terms of the sequence have $\nu_p$ non-negative so at least $n-\left\lceil \frac{n}{p^k} \right\rceil$ terms have $\nu_p$ negative which means that for any such $\frac{a_i}{b_i}$ we must have $p \mid b_i$, now since $b_i \le 5n$ we have that there's at most $\left\lceil \frac{5n}{p} \right\rceil$ multiples of $p$ among all the $b$'s which gives: $$n-\frac{n}{p^k}-1 \le n-\left\lceil \frac{n}{p^k} \right\rceil \le \left\lceil \frac{5n}{p} \right\rceil \le \left\lceil \frac{5n}{p^k} \right\rceil \le \frac{5n}{p^k}+1 \implies n-2 \le \frac{6n}{p^k} \le \frac{6n}{7} \implies n \le 14$$The last thing is clearly a contradiction, but this works for $k=1$ and $p \ge 7$, now otherwise use the lemma on $\ell=1$ then we have that at least $n-\left\lceil \frac{n}{p} \right\rceil$ terms have their $\nu_p$ on the minimun, but there's at most $\left\lceil \frac{5n}{p^k} \right\rceil$ such $b$ terms, therefore if $k \ge 2$ for $p \ge 3$ we obtain our desired contradiction. In addition for $p=2$ we have that $\nu_p \left(\frac{a_i}{b_i} \right) \ge -3$ for all $1 \le i \le n$ using the same method as above. Note that if there was ever a term with $\nu_2 \ge -2$ then we would also get that $\nu_2(r) \ge -2$ for $r$ being the difference in the sequence. So now if we had that $\nu_2 =-3$ for all terms in the arithmetic sequence and even $\nu_2(r)=-3$, it would turn out that we need $n$ multiples of $8$ but we can only get at most $\left\lceil \frac{5n}{8} \right\rceil$ which is a contradiction for $n \ge 2018$, therefore either way $\nu_2(r) \ge -2$ The finish: A direct consequence from our claim and the extra work is that $60r \in \mathbb Z$, now WLOG the sequence is strictly increasing (assume FTSOC $r>0$) which means that $r \ge \frac{1}{60}$, now among any consecutive $\left\lfloor \frac{n}{2} \right\rfloor+1$ terms in the $b_i$'s we have that one of them is greater or equal to $\left\lfloor \frac{n}{2} \right\rfloor+1$, so we pick one such from $b_{\left\lceil \frac{n}{2} \right \rceil}, \cdots, b_n$ and just note that: $$\frac{1}{60} \le r=\frac{\frac{a_i}{b_i}-\frac{a_1}{b_1}}{i-1}<\frac{a_i}{b_i(i-1)} \le \frac{5n}{\left(\left\lfloor \frac{n}{2} \right\rfloor+1 \right) \cdot \left(\left\lceil \frac{n}{2} \right\rceil-1 \right)} \le \frac{5n}{\left\lceil \frac{n}{2} \right\rceil \cdot \left\lfloor \frac{n}{2} \right\rfloor-1} \implies \left\lceil \frac{n}{2} \right\rceil \cdot \left\lfloor \frac{n}{2} \right\rfloor-1<300n \implies \left(\left\lceil \frac{n}{2} \right\rceil-300 \right) \left(\left\lfloor \frac{n}{2} \right\rfloor-300 \right) \le 300^2$$$$714^2 \le \left(\left\lceil \frac{n}{2} \right\rceil-300 \right) \left(\left\lfloor \frac{n}{2} \right\rfloor-300 \right) \le 300^2 \; \text{contradiction!}$$Therefore it turns out that $r=0$ so the sequence is constant thus we are done .

Insanely cool problem Let the common difference be $\frac{a}{b}$ Claim1: All the indices such that $\nu_{p}(\frac{a_i}{b_i}) < \nu_{p}(\frac{a}{b})$ are congruent modulo p.\ Proof: FTCOS assume that there exists $i,j$ such that $\nu_{p}(b_i) < \nu_{p}({b})$ and $\nu_{p}({b_j}) < \nu_{p}({b})$ and $i \not\equiv j$ mod p. Then we get that \[\nu_{p}((i-j)\frac{a}{b}) = \nu_{p}(\frac{a_i}{b_i} - \frac{a_j}{b_j}) \geq \min\{ \nu_{p}(\frac{a_i}{b_i}),\nu_{p}(\frac{a_j}{b_j})\} \]But this works only when $i \equiv j$ mod p as the least common denominator is not divisible by $p^{\nu_p(b)}$ contradiction. $\square$ Claim 2: The prime factors of $b$ are at most 5. Proof: Claim 1 shows that at most $\lceil \frac{n}{p}\rceil \leq \frac{n}{p}+1$ such vertices are there that have vp(b) > vp(b_i). so at least $(\frac{p-1}{p})n -1$ does not satisfy the given condition. Then assume contrary and we get \[ 5n \geq \max\{b_i :p|b_i\} \geq ((\frac{p-1}{p})n -1)p = (p-1)(n-1) -1 \geq 6(n-1)-1 > 5n\] Contradiction. $\square$ Claim 3: $b \geq \frac{n-2}{20}$ Proof: Remove all $b_i$ such that $b_i < \frac{n}{2}$ then the rest must at most $\frac{5n}{\frac{n}{2}} =10$, then this just gives us that $b>\frac{n-2}{20}>19$ done Now the final step is: \[5n > \max\{b_i: d|b_i\} > (\lceil \frac{n}{30} \rceil \cdot 8)\cdot b > (\lceil \frac{n}{30}\rceil \cdot 8)\cdot \frac{n-2}{20} > 5n\] Contradiction so we are done. PS: this is actually my first N7

EthanWYX2009 wrote: $5n$ can be improved to $\mathcal O\left(\sqrt{\frac{n^3}{\ln\ln n}}\right).$ May I get the source or proof of this statement?