Problem 1384

1. Claim: Let $d(n)$ be the divisor-counting function. Then $d(n) = O(n^k)$ for all $k > 0$. Here, we use the actual definition of big-O, i.e., $f(n) = O(g(n)) \iff |f(n)| \leq Cg(n)$ for some fixed constant $C$.

2. Proof of Claim:
- For some prime power $n = p^e$, we have $d(n) = e + 1$.
- Consider the ratio $\frac{d(n)}{n^k} = \frac{e + 1}{p^{e/k}}$.
- There exists a constant $M$ (dependent on $k$) such that for any $p > M$, this fraction will always be at most 1.
- Therefore, $\frac{d(n)}{n^k}$ is bounded above by
$$\prod_{\substack{p \text{ prime} \\ p \leq M}} \max \left\{ \frac{e + 1}{p^{e/k}} \right\}_{e \geq 0}$$
which is clearly finite, proving the desired claim. $\blacksquare$

3. Now, pick $k = \frac{1}{1000}$ for our lemma. By picking $n$ as shown in the problem, we have $d(n) = O(n^{1/1000})$.

4. There are $O(n^{1/500})$ pairs of distinct divisors $(d_1, d_2)$ of $n$.

5. For one of these pairs, there are exactly $d(|d_1 - d_2|)$ positive numbers $m$ such that $d_1$ and $d_2$ are not distinct modulo $m$.

6. By the union bound, this means that the number of $m$ such that the elements of $D_n$ aren't pairwise distinct modulo $m$ is at most
$$\sum_{(d_1, d_2)} d(|d_1 - d_2|) \leq \sum_{(d_1, d_2)} d(n) = \sum_{(d_1, d_2)} O(n^{1/1000}) = O(n^{3/1000}),$$
which will clearly be less than $n^{1/100}$ for sufficiently large $n$. $\blacksquare$