Problem 1700

1. Claim: There exist infinitely many positive integers $a$ such that the equation $\tau(an) = n$ does not have a positive integer solution $n$.

2. Consideration: Let $a = p^{14}$ for a sufficiently large prime $p$.

3. Prime Factorization: Suppose $n = p^f p_1^{f_1} \ldots p_k^{f_k}$ for positive integers $f_i$. Then we need:
$$\tau(an) = \tau(p^{14} n) = \tau(p^{14} p^f p_1^{f_1} \ldots p_k^{f_k}) = (14 + f + 1)(f_1 + 1) \ldots (f_k + 1) = n = p^f p_1^{f_1} \ldots p_k^{f_k}$$

4. **Applying $\omega$**: Let $\omega(m)$ denote the number of prime powers greater than 1 dividing $m$. Note that $\omega$ is completely multiplicative. Applying it to both sides of the equation, we get:
$$\omega(15 + f) + \omega(f_1 + 1) + \cdots + \omega(f_k + 1) = f + f_1 + \cdots + f_k$$
This implies:
$$(f_1 - \omega(f_1 + 1)) + \cdots + (f_k - \omega(f_k + 1)) = \omega(15 + f) - f$$

5. **Bounding $\omega$**: Since $\omega(m) \leq \tau(m) \leq 2\sqrt{m}$, it follows that for fixed positive integers $r$, $\omega(m + r) - m$ is bounded above and eventually negative as $m$ varies. Furthermore, we can check (by examining small $m$ manually) that we must always have $m - \omega(m + 1) \geq 0$.

6. **Bounding $f_i$**: In particular, this means that, by appealing to the fact that $\omega(15 + f) - f$ is bounded above, $f_i - \omega(f_i + 1)$ is bounded above as well, hence the $f_i$ are all bounded above. Furthermore, since $f - \omega(15 + f)$ should be nonpositive, $f$ is bounded above as well. Hence if $p$ is sufficiently large, we are forced to have $f = 0$, else $p$ divides $n$ but $p \nmid (15 + f)(f_1 + 1) \ldots (f_k + 1)$ for size reasons.

7. **Casework on $f_i$**: Hence the RHS of the equation just equals 2, forcing $f_i - \omega(f_i + 1) \leq 2$ for all $i$. We may manually verify that for positive integers $m$, $m - \omega(m + 1) \leq 2 \implies m \in \{1, 2, 3\}$, hence $f_i \in \{1, 2, 3\}$ for all $i$. Now we do casework.

8. **Casework on $\nu_5(n)$**:
- Because $15 \mid n$, we should have $\nu_5(n) \in \{1, 2, 3\}$. On the other hand, if $\nu_5(n) \geq 2$ then we need some prime $q$ to have $\nu_q(n) \equiv 4 \pmod{5}$, but this contradicts the fact that it should be at most 3 always. Thus $\nu_5(n) = 1$.

9. **Casework on $\nu_3(n)$**:
- If $\nu_3(n) = 1$, then $\nu_2(n) \in \{2, 3\}$, since now $(1 + 1)(1 + 1) \mid \tau(an)$. But $\nu_2(n) = 2$ would imply $9 \mid \tau(an)$ and $\nu_2(n) = 3$ would imply $16 \mid \tau(an)$, so this case is impossible.
- If $\nu_3(n) = 2$, then $\nu_2(n) \in \{1, 2, 3\}$. But $\nu_2(n) = 1$ would imply $4 \mid \tau(an)$: contradiction. On the other hand, if $\nu_2(n) \in \{2, 3\}$, we are faced with a problem of having "too many" factors of 2, so there has to be some prime $q \mid n$ which we haven't considered yet that has odd exponent. But then $q \geq 7$, so $q \nmid 15$ and $q \nmid f_i + 1$ due to size reasons: also a contradiction.
- If $\nu_3(n) = 3$, then $\nu_2(n) = 3$ is forced since $8 \mid \tau(an)$. But then $32 \mid \tau(an)$, which is again impossible.

10. Conclusion: This finishes the problem. $\blacksquare$