Subfield: Number Theory


Question: Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5 n$. Suppose that the sequence

$$
\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}}
\tag{1}
$$

forms an arithmetic progression. Prove that the terms of the sequence are equal.

Solution:
Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\Delta=\frac{c}{d}$, where $d>0$ and $c, d$ are coprime.

We will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \leqslant i \leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\left(b_{i}\right)<v_{p}(d) .\left(v_{p}(x)\right.$ stands for the exponent of $p$ in the prime factorisation of $x$.)

Claim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$.

Proof. Let $\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $\left.p^{\alpha}\right)$ such that $i \not \equiv j(\bmod p)$. Then the least common denominator of $\frac{a_{i}}{b_{i}}$ and $\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\alpha}$. But this is impossible because in their difference, $(i-j) \Delta=\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\alpha}$ divides the denominator $d$.

Claim 2. $d$ has no prime divisors greater than 5 .

Proof. Suppose that $p \geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \ldots, n$, at most $\left\lceil\frac{n}{p}\right\rceil<\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\frac{p-1}{p} n-1$ of $b_{1}, \ldots, b_{n}$. Since these denominators are distinct,

$$
5 n \geqslant \max \left\{b_{i}: p \mid b_{i}\right\} \geqslant\left(\frac{p-1}{p} n-1\right) p=(p-1)(n-1)-1 \geqslant 6(n-1)-1>5 n
$$

a contradiction.

Claim 3. For every $0 \leqslant k \leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \ldots, b_{k+30}$, at least $\varphi(30)=8$ are divisible by $d$.

Proof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \cdot(3-1) \cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \mid b_{i}$ for every uncovered index $i$.

Claim 4. $|\Delta|<\frac{20}{n-2}$ and $d>\frac{n-2}{20}$.

Proof. From the sequence (1), remove all fractions with $b_{n}<\frac{n}{2}$, There remain at least $\frac{n}{2}$ fractions, and they cannot exceed $\frac{5 n}{n / 2}=10$. So we have at least $\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\frac{10}{n / 2-1}=\frac{20}{n-2}$.

The second inequality follows from $\frac{1}{d} \leqslant \frac{|c|}{d}=|\Delta|$.

Now we have everything to get the final contradiction. By Claim 3, we have $d \mid b_{i}$ for at least $\left\lfloor\frac{n}{30}\right\rfloor \cdot 8$ indices $i$. By Claim 4, we have $d \geqslant \frac{n-2}{20}$. Therefore,

$$
5 n \geqslant \max \left\{b_{i}: d \mid b_{i}\right\} \geqslant\left(\left\lfloor\frac{n}{30}\right\rfloor \cdot 8\right) \cdot d>\left(\frac{n}{30}-1\right) \cdot 8 \cdot \frac{n-2}{20}>5 n .
$$