Question 98
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 $d\left(a^{2}+b^{2}\right)$ for any positive integers $a, b$ satisfying $a+b=n$.