  We prove that $$ \sum_{p \leq x} \frac{1}{\tau(p-1)} \asymp \frac{x}{(\log
x)^{3/2}}, \quad \quad \sum_{n \leq x} \frac{1}{\tau(n^2+1)} \asymp
\frac{x}{(\log x)^{1/2}}, $$ where $\tau(n)=\sum_{d|n}1$ is the number of
divisors of $n$, and the summation in the first sum is over primes.

Number Theory