Problem: We denote the number of positive divisors of a positive integer $m$ by $d(m)$
and the number of distinct prime divisors of $m$ by $\omega(m)$.
Let $k$ be a positive integer.
Prove that there exist infinitely many positive integers $n$
such that $\omega(n) = k$ and
$d(n)$ does not divide $d(a^2+b^2)$
for any positive integers $a, b$ satisfying $a + b = n$.

Solution: Let $n = 2^{p-1}t$, where $t \equiv 5 \pmod 6$, $\omega(t) = k-1$,
and $p \gg t$ is a sufficiently large prime.
Let $a+b=n$ and $a^2+b^2=c$.
We claim that $p \nmid d(c)$,
which solves the problem since $p \mid d(n)$.

First, note that $3 \nmid a^2+b^2$, since $3 \nmid n$.
Next, note that $c < 2n^2 < 5^{p-1}$ (since $p \gg t$)
so no exponent of an odd prime in $c$ exceeds $p-2$.
Moreover, $c < 2^{3p-1}$.

So, it remains to check that $\nu_2(c) \notin \{p-1, 2p-1\}$.
On the one hand, if $\nu_2(a) < \nu_2(b)$,
then $\nu_2(a) = p-1$ and $\nu_2(c) = 2\nu_2(a) = 2p-2$.
On the other hand, if $\nu_2(a) = \nu_2(b)$
then $\nu_2(a) \le p-2$, and $\nu_2(c) = 2\nu_2(a)+1$
is odd and less than $2p-1$.

\begin{remark*}
  Personally, I find the condition to be artificial,
  but the construction is kind of fun.

  I also think the scores on this problem during the real contest
  are low mostly because of the difficulty of problem 2.
\end{remark*}