Originally Posted by
melese Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$\displaystyle m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $\displaystyle n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$, where for $\displaystyle 1\leq i\leq r$ $\displaystyle m_i$ and $\displaystyle n_i$ are nonnegative. Also, for any prime $\displaystyle p_i$, $\displaystyle p_i$ divides at leat one of $\displaystyle m$ and $\displaystyle n$.
Now, $\displaystyle mn=p_1^{m_1+n_1}p_2^{m_2+n_2}\dots p_r^{m_r+n_r}$ and then $\displaystyle \tau(mn)=(m_1+n_1+1)(m_2+n_2+1)\dots (m_r+n_r+1)$.
Computing $\displaystyle \tau(m)\tau(n)$ gives $\displaystyle \tau(m)\tau(n)=(m_1+1)(n_1+1)\dots... (m_r+1)(n_r+1)$ $\displaystyle =(m_1n_1+m_1+n_1+1)\dots (m_rn_r+m_r+n_r+1)$.
Now comes my difficulty...
For each $\displaystyle i=1,2,\dots ,r$, $\displaystyle m_in_i+m_i+n_i+1\leq m_i+n_i+1$, so if we want $\displaystyle \tau(mn)=\tau(m)\tau(n) $ we must have
$\displaystyle m_in_i+m_i+n_i+1= m_i+n_i+1$ for $\displaystyle i=1,2,\dots ,r$. Otherwise $\displaystyle \tau(mn)<\tau(m)\tau(n)$.
But then $\displaystyle m_in_i=0$. Without loss of generality $\displaystyle m_i=0$ and this means that $\displaystyle p_i$ divides n but not m. In this manner $\displaystyle m$ and $\displaystyle n$ are relatively prime.
By the way exactly one of $\displaystyle m_i$ and $\displaystyle n_i$ equals $\displaystyle 0$ due to: "for any prime $\displaystyle p_i$, $\displaystyle p_i$divides at leat one of $\displaystyle m$ and $\displaystyle n$."
Is my argument right? Thanks for your help.