Is the converse statement true?

**melese** · June 1st 2010, 06:18 PM

Let $m,n$ be natural numbers.
Does $\tau(mn)=\tau(m)\tau(n)$
, where $\tau(k)$ is number of positive divisors of k, implies that
$\gcd(m,n)=1$ ?

Thanks.

**chiph588@** · June 1st 2010, 06:27 PM

Originally Posted by melese

Let $m,n$ be natural numbers.
Does $\tau(mn)=mn$
, where $\tau(k)$ is number of positive divisors of k, implies that
$\gcd(m,n)=1$ ?

Thanks.

Since your title says converse, I assume you meant to say $\tau(mn)=\tau(m)\tau(n)$ ?

**Bruno J.** · June 1st 2010, 06:29 PM

You probably mean $\tau(mn)=\tau(m)\tau(n)$ ?

**Bruno J.** · June 1st 2010, 06:30 PM

Originally Posted by chiph588@

Since your title says converse, I assume you meant to say $\tau(mn)=\tau(m)\tau(n)$ ?

Too fast

**chiph588@** · June 1st 2010, 06:46 PM

Originally Posted by Bruno J.

Too fast

Too slow

Originally Posted by melese

Let $m,n$ be natural numbers.
Does $\tau(mn)=mn$
, where $\tau(k)$ is number of positive divisors of k, implies that
$\gcd(m,n)=1$ ?

Thanks.

Correct me if I'm wrong anyone but I think it might be true...

Here's some reasoning:

Suppose $mn=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ and $m=p_1^{a_1}p_2^{a_2}\cdots p_i^{b_i}\cdots p_j^{b_j}$ and $n=p_i^{a_i-b_i}\cdots p_j^{a_j-b_j}p_{j+1}^{a_{j+1}}\cdots p_k^{a_k}$ .

$\tau(mn)=\prod_{s=1}^k (a_s+1) = \tau(m)\tau(n)$

Omitting the simplification we get $\prod_{s=i}^j (a_s+1)=\prod_{s=i}^j (b_s+1)(a_s-b_s+1) = \prod_{s=i}^j (b_s(a_s-b_s)+a_s+1)$

But $b_r>0\implies a_r+1<b_r(a_r-b_r)+a_s+1$ , so we're forced to have $b_s=0 \; \forall \; s$ if we want this equality to hold.

**Bruno J.** · June 1st 2010, 08:27 PM

If you prove it, then it's true!

Here's a less computational argument. Let $D_n$ denote the set of divisors of $n$ , and suppose that $g=(m,n)>1$ . I exhibit a surjection $f : \ D_m \times D_n \to D_{mn}$ which is not an injection. For every $(u,v) \in D_m \times D_n$ , let $f(u,v)=uv$ . This map is clearly a surjection. But $f(m/g,n)=f(m,n/g)$ . Therefore $|D_m\times D_n|>|D_{mn}|$ .

**melese** · June 2nd 2010, 06:33 AM

yes I do, I wrote that early in the morning...

**melese** · June 2nd 2010, 06:44 AM

Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$ , where for $1\leq i\leq r$ $m_i$ and $n_i$ are nonnegative. Also, for any prime $p_i$ , $p_i$ divides at leat one of $m$ and $n$ .

Now, $mn=p_1^{m_1+n_1}p_2^{m_2+n_2}\dots p_r^{m_r+n_r}$ and then $\tau(mn)=(m_1+n_1+1)(m_2+n_2+1)\dots (m_r+n_r+1)$ .
Computing $\tau(m)\tau(n)$ gives $\tau(m)\tau(n)=(m_1+1)(n_1+1)\dots... (m_r+1)(n_r+1)$ $=(m_1n_1+m_1+n_1+1)\dots (m_rn_r+m_r+n_r+1)$ .

Now comes my difficulty...
For each $i=1,2,\dots ,r$ , $m_in_i+m_i+n_i+1\leq m_i+n_i+1$ , so if we want $\tau(mn)=\tau(m)\tau(n)$ we must have
$m_in_i+m_i+n_i+1= m_i+n_i+1$ for $i=1,2,\dots ,r$ . Otherwise $\tau(mn)<\tau(m)\tau(n)$ .

But then $m_in_i=0$ . Without loss of generality $m_i=0$ and this means that $p_i$ divides n but not m. In this manner $m$ and $n$ are relatively prime.
By the way exactly one of $m_i$ and $n_i$ equals $0$ due to: "for any prime $p_i$ , $p_i$ divides at leat one of $m$ and $n$ ."

Is my argument right? Thanks for your help.

**chiph588@** · June 2nd 2010, 10:50 AM

Originally Posted by melese

Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$ , where for $1\leq i\leq r$ $m_i$ and $n_i$ are nonnegative. Also, for any prime $p_i$ , $p_i$ divides at leat one of $m$ and $n$ .

Now, $mn=p_1^{m_1+n_1}p_2^{m_2+n_2}\dots p_r^{m_r+n_r}$ and then $\tau(mn)=(m_1+n_1+1)(m_2+n_2+1)\dots (m_r+n_r+1)$ .
Computing $\tau(m)\tau(n)$ gives $\tau(m)\tau(n)=(m_1+1)(n_1+1)\dots... (m_r+1)(n_r+1)$ $=(m_1n_1+m_1+n_1+1)\dots (m_rn_r+m_r+n_r+1)$ .

Now comes my difficulty...
For each $i=1,2,\dots ,r$ , $m_in_i+m_i+n_i+1\leq m_i+n_i+1$ , so if we want $\tau(mn)=\tau(m)\tau(n)$ we must have
$m_in_i+m_i+n_i+1= m_i+n_i+1$ for $i=1,2,\dots ,r$ . Otherwise $\tau(mn)<\tau(m)\tau(n)$ .

But then $m_in_i=0$ . Without loss of generality $m_i=0$ and this means that $p_i$ divides n but not m. In this manner $m$ and $n$ are relatively prime.
By the way exactly one of $m_i$ and $n_i$ equals $0$ due to: "for any prime $p_i$ , $p_i$ divides at leat one of $m$ and $n$ ."

Is my argument right? Thanks for your help.

You mean $m_in_i+m_i+n_i+1\geq m_i+n_i+1$ .

Other than that, it looks fine to me.

**melese** · June 3rd 2010, 05:23 AM

Again, you corrected me. Thank you so much for your help!

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