SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) Denote $d(x)$, $\pi (x)$ as the number of divisors, the product of divisors of positive integer $x$. Firstly, we can see that for any divisor $y_k$ of $x$, $1\le k\le d(x)$ then $\frac{x}{y_i}$ is also divisor of $x$, thus $$\prod _{k=1}^{d(x)}{y}_k=\prod _{k=1}^{d(x)}\frac{x}{y_k}\text{ so }{\left(\prod _{k=1}^{d(x)}{y}_k\right)}^2=x^{d(x)},\text{ hence }\pi (x)=x^{\frac{d(x)}{2}}$$ Thus for any $m,n\in {\mathbb{Z}}^{+}$ then $\pi (m)=\pi (n)$ implies that $m,n$ share the common prime divisors set $S$. Suppose that $d(m)\ge d(n)$ and take $p\in S$. Since $m^{d(m)}=n^{d(n)}$, we have $d(m)\cdot v_p(m)=d(n)\cdot v_p(n)$. Since $d(m)\ge d(n)$, we get $v_p(m)\le v_p(n)$. And this is true for all $p\in S$, thus $$d(m)=\prod _{p\in S}(v_p(m)+1)\le \prod _{p\in S}(v_p(n)+1)=d(n)$$ So $d(m)=d(n)$, which implies that $m=n$. Therefore, we can find a positive integer such that product of all divisors of $sm$, $sn$ are equal if and only if $m=n$.

2) Firstly, we can see that if $n\mid m$ then any divisor of $sn$ is also divisor of $sm$, so $d(sn)<d(sm)$. We consider $n\nmid m$, and denote $p_1,p_2,\dots ,p_t$ be all prime dividing $mn$. Suppose that $$m=\prod _{i=1}^t{p}_i^{{\alpha }_i}\text{ and }n=\prod _{i=1}^t{p}_i^{{\beta }_i}$$ Now we are looking for $s={\prod }_{i=1}^t{p}_i^{{\gamma }_i}$ such that $$\frac{d(sm)}{d(sn)}=\prod _{i=1}^t\frac{{\alpha }_i+{\gamma }_i+1}{{\beta }_i+{\gamma }_i+1}=1$$ Note that if ${\alpha }_i={\beta }_i$, then regardless of the value of ${\gamma }_i$, the corresponding factor equals to 1 and does not affect the product. So we may assume that ${\alpha }_i\ne {\beta }_i$ for all $1\le i\le t$.

Claim. Let $\alpha >\beta$ be nonnegative integers. Then for every $M\ge \beta +1$, there exist a nonnegative integer $\gamma$ such that $$\frac{\alpha +\gamma +1}{\beta +\gamma +1}=\frac{M+1}{M}.$$ It is equivalent to $\gamma =M(\alpha -\beta )-(\beta +1)\ge 0$, which is true.

Back to the original problem, we can assume that ${\alpha }_i>{\beta }_i$ for $i=1,2,\dots ,u$ and ${\alpha }_i<{\beta }_i$ for $i=u+1,u+2,\dots ,t$. Take some big enough $X$ and choose ${\gamma }_i$ such that - $\frac{{\alpha }_i+{\gamma }_i+1}{{\beta }_i+{\gamma }_i+1}=\frac{uX+i}{uX+i-1}$ for $1\le i\le u$. - $\frac{{\beta }_{u+i}+{\gamma }_{u+i}+1}{{\alpha }_{u+i}+{\gamma }_{u+i}+1}=\frac{(t-u)X+i}{(t-u)X+i-1}$ for $1\le i\le t-u$. Then we have $$\frac{d(sm)}{d(sn)}=\prod _{i=1}^u\frac{uX+i}{uX+i-1}\cdot \prod _{i=1}^{t-u}\frac{(t-u)X+i-1}{(t-u)X+i}=\frac{u(X+1)}{uX}\cdot \frac{(t-u)X}{(t-u)(X+1)}=1.$$