Perhaps it was posted before. Solution Yes. They are equal. Proof Suppose that $ \prod _{d|m} d= \prod _{d|n} d $ $\quad$for some$ ( m,n) \in \mathbb{N}^{*} \times \mathbb{N}^{*}$ Then we have $ m^{\tau ( m) /2} =n^{\tau ( n) /2}$, which implies $ m^{\tau ( m)} =n^{\tau ( n)}$... $ ( \heartsuit )$ So it is enough to show that $ \tau ( m) =\tau ( n)$. Without loss of generality, assume that $ \tau ( m) \geq \tau ( n)$ ...$ ( \spadesuit )$ Let $ p$ be an arbitrary prime number. From $ ( \heartsuit )$, we have $ \tau ( m) \cdotp \nu _{p}( m) =\tau ( n) \cdotp \nu _{p}( n)$. From this relation and $ ( \spadesuit )$, we have $ \tau ( n) \cdotp \nu _{p}( n) \geq \tau ( n) \cdotp \nu _{p}( m)$. It follows that $ \nu _{p}( n) \geq \nu _{p}( m)$, which implies $ m\mid n$. Thus, in particular, we have $ \tau ( n) \geq \tau ( m)$. From this relation and $ ( \spadesuit )$, we must have $ \tau ( m) =\tau ( n)$. $ \blacksquare $