Quote: Determine all pairs $(m, n)$ of positive integers for which there exists a positive integer $s$ such that $sm$ and $sn$ have an equal number of divisors. Assume henceforth that $m \ne n$ or it's immediate (take any $s$). Obviously if $m \mid n$ or $n \mid m$ then no such $s$ exists. We will prove that conversely that for all other $(m,n)$ such $s$ exists. There are many constructions. It seems cleanest to start by saying: Claim: If $\alpha > \beta \ge 0$ are integers then for any $M \ge \beta+1$ there exists $\gamma \in {\mathbb Z}_{\ge 0}$ such that \[ \frac{\alpha+\gamma+1}{\beta+\gamma+1} = \frac{M+1}{M}. \]The claim is checked by setting $\gamma = M(\alpha-\beta) - \beta+1$. Now, suppose for primes we write \begin{align*} m &= p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k} q_1^{\alpha'_1} \dots q_\ell^{\alpha'_\ell} \\ n &= p_1^{\beta_1} p_2^{\beta_2} \dots p_k^{\beta_k} q_1^{\beta'_1} \dots q_\ell^{\beta'_\ell} \end{align*}where $\alpha_i > \beta_i$ and $\alpha_j' < \beta_j'$; by hypothesis $\min(k,\ell) \ge 1$. We have here omitted any primes $p$ for which $\nu_p(m) = \nu_p(n)$, since these do not affect the proof in any way. Let $T$ be a large integer exceeding $\max(\alpha_i, \beta_j)$. We can pick $\gamma_i$ for $i=1,\dots,k$ such that \[ \prod_i \frac{\gamma_i + \alpha_i + 1}{\gamma_i + \beta_i + 1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \]Similarly we can pick $\gamma_i'$ for $i=1,\dots,\ell$ such that \[ \prod_j \frac{\gamma'_j + \beta'_j + 1}{\gamma'_j + \alpha'_j + 1} = \frac{\ell T+1}{\ell T} \cdot \frac{\ell T+2}{\ell T+1} \cdot \dots \cdot \frac{\ell T+\ell }{\ell T+(\ell -1)} = \frac{\ell T+\ell }{\ell T} = \frac{T+1}{T}. \]Then if we let $s = \prod_i p_i^{\gamma_i} \prod_i q_i^{\gamma'_i}$, we have \[ \frac{d(sm)}{d(sn)} = \prod_i \frac{\gamma_i + \alpha_i + 1}{\gamma_i + \beta_i + 12} \prod_j \frac{\gamma'_j + \alpha'_j + 1}{\gamma'_j + \beta'_j + 1} = \frac{T+1}{T} \cdot \frac{T}{T+1} = 1 \]as desired.

Use $ka$ and $kb$ instead of $sn$ and $sk$. The answer is all pairs of integers such that $a\nmid b$ and $b\nmid a$. First, if $a\mid b$, then any divisor of $ka$ is also divisor of $kb$ but since $a\neq b$, we easily obtain that $\tau(ka) <\tau(kb)$ for any $k$ so this pair does not work. It remains to give construction to any other pair. Since it's clear that we can cancel any prime factor which have the same exponents on $a,b$, we can write $$ a = \prod_{i=1}^m p_i^{a_i-1} \prod_{j=1}^n q_j^{c_j-1}, \qquad b = \prod_{i=1}^m p_i^{b_i-1} \prod_{j=1}^n q_j^{d_j-1} $$where $a_1,...,a_m, b_1,...,b_m,c_1,...,c_n,d_1,...,d_n$ are integers greater than one such that $a_i > b_i$ and $c_i > d_i$, and $p_1,...,p_m,q_1,...,q_m$ are primes. Take a large enough integer $T$ and define \begin{align*} x_i &= (mT+i)(a_i-b_i)-a_i\\\ y_j &= (nT+j)(d_j-c_j)-d_j. \end{align*}Choose $n=\prod\limits_{i=1}^m p_i^{x_i-1} \prod\limits_{j=1}^m q_j^{y_j-1}$. We see that $$\frac{x_i+a_i}{x_i+b_i}=\frac{mT+i}{mT+i-1},\qquad \frac{y_j+c_j}{y_j+d_j}=\frac{nT+j-1}{nT+j}$$Hence $$\frac{\tau(na)}{\tau(nb)} = \prod_{i=1}^m\frac{mT+i}{mT+i-1}\prod_{j=1}^n\frac{nT+j-1}{nT+j} = \frac{m(T+1)}{mT}\cdot\frac{nT}{n(T+1)}=1$$as desired.

I claim that all $(m,n)$ such that $m\!\!\not\arrowvert n$, $n\!\!\not\arrowvert m$ work. Suppose that $(p_1,p_2,\ldots,p_k)$ are the set of primes such that $p|\text{lcm}(m,n)$, and define $v_{p_i}(m)=\alpha_i$, $v_{p_i}(n)=\beta_i$. Finally, order the $p_i$ such that for all $i< N$, $\alpha_i<\beta_i$, and $\alpha_i> \beta_i$* for $i\ge N$. $2\le N\le k$ by our assumption that neither number divides the other. Now, if we denote $v_{p_i}(s)=s_i$, we want to find a suitable choice of $(s_1,\ldots,s_k)$ such that $$\prod_{i=1}^k\frac{\alpha_i+s_i}{\beta_i+s_i}=1$$Call the factor inside the product $F_i$. Note that for all $i\ge N$, we can choose $s_i=T\alpha_i-(T+1)\beta_i$, for all sufficiently large $T$, such that $F_i=\frac{T+1}{T}$. Similarly, for all $i<N$, we can choose $s_i$ such that $F_i=\frac{T}{T+1}$, for all sufficiently large $T$. So, we can set $F_1=\frac{T_1}{T_1+1}$, $F_2=\frac{T_1+1}{T_1+2}\ldots F_{N-1}=\frac{T_1+N-2}{T_1+N-1}$ for some very large $T_1$, and $F_N=\frac{T_2+1}{T_2},\ldots F_k=\frac{T_2+k-N+1}{T_2+k-N}$ for some very large $T_2$. Now, the desired product telescopes and becomes $\frac{T_1}{T_1+N-1}\cdot\frac{T_2+(k-N+1)}{T_2}$. Now, just let $T_1= M(N-1)$, $T_2=M(k-N+1)$ for some very large value of $M$ to get this equal to 1, as desired. *If there are any $p$ such that $v_p(m)=v_p(n)$, we can divide them out since they don't matter.

This solution might be a bit new. We claim the only condition is $m\nmid n$ and $n\nmid m$. Clearly $m=n$ works, so assume $m\not =n$ in what follows. Suppose $m = \prod_{i=1}^k p_i^{a_i}, n = \prod_{i=1}^k p_i^{b_i}, s = \prod_{i=1}^k p_i^{x_i}$. (We ignore the primes that divide equally in $m,n$.) Then \[ (x_1+(a_1+1))\cdots (x_k + (a_k+1)) = (x_1+(b_1+1))\cdots (x_k + (b_k+1)). \]If $a_1\le b_i \forall 1\le i \le k$ or $a_1\ge b_i \forall 1\le i \le k$, then clearly one side is larger than the other, so we have no solutions. This corresponds to $m\mid n$ or $n\mid m$. So assume this is not the case. WLOG assume $a_i\le b_i$ for $i=1,\ldots,\ell$ and $a_i \ge b_i$ for $i=\ell+1,\ldots, k$. We can do this since the value of the primes is irrelevant, only their exponents, so we can order the indices $i$ with $a_i\le b_i$ first. Then \[ \frac{x_1+b_1+1}{x_1+a_1+1} \cdot \frac{x_2+b_2+1}{x_2+a_2+1} \cdots \frac{x_\ell+b_\ell+1}{x_\ell+a_\ell+1} = \frac{x_{\ell+1}+b_{\ell+1}+1}{x_{\ell+1}+a_{\ell+1}+1} \cdots \frac{x_k+b_k+1}{x_k+a_k+1}. \]We want to find a solution for the $x_i$'s. Let \begin{align*} x_1+a_1+1 &= x_2+b_2+1 \implies x_2 = x_1+(a_1-b_2) \\ x_2+a_2+1 &= x_3+b_3+1 \implies x_3 = x_2+(a_2-b_3) \\ &\vdots \\ x_{\ell-1}+a_{\ell-1} + 1 &= x_\ell + b_\ell + 1\implies x_\ell = x_{\ell-1} + (a_{\ell-1} - b_\ell), \end{align*}and similarly, \begin{align*} x_{\ell+1} + b_{\ell+1} + 1 &= x_{\ell+2} + a_{\ell+2} + 1 \\ &\vdots \\ x_{k-1} + b_{k-1} + 1&= x_k+a_k+1. \end{align*}Now most of the fractions cancel, and we are left with \[ \frac{x_1+b_1+1}{x_\ell+a_\ell+1} = \frac{x_{\ell+1} + a_{\ell+1} + 1}{x_k+b_k+1}. \]This clearly has a solution for $x_1, x_\ell, x_{\ell+1}, x_k$; just take $x_1 = A-(b_1+1), x_{\ell+1} = A-(a_{\ell+1}+1), x_{\ell} = B-(a_{\ell}+1), x_k = B-(b_k+1)$, for some very large $A,B$. Now we just substitute into the above equations to find all $x_i$'s. Note that if we make $A,B$ sufficiently large then we guarantee that all the $x_i$'s are positive.

darn just realized that this is basically the same (rather basically identical) as pad's solution but im posting it anyways for reference

IndoMathXdZ wrote: Determine all pairs $(m, n)$ of positive integers for which there exists a positive integer $s$ such that $sm$ and $sn$ have an equal number of divisors. djmathman's edit reason wrote: edited to reflect actual wording in the shortlist re post #2 It's quite ironic that the text in post 2 is not the actual shortlist wording. Fixed, thanks. ~dj

I think this is slightly different? The answer is all $(k,n)$ such that neither divides the other. It's easy to see that if $k|n$ or $n|k$, then it doesn't work. Now we show that all other $(k,n)$ work. Let \begin{align*} n &=p_1^{d_1}p_2^{d_2} \cdots p_t^{d_t} \\ k &=p_1^{e_1}p_2^{e_2} \cdots p_t^{e_t} \\ s &=p_1^{f_1-1}p_2^{f_2-1} \cdots p_t^{f_t-1} \\ \end{align*}for primes $p_i$, positive integers $f_i$ and nonnegative integers $d_i, e_i$. We can WLOG let $d_i \neq e_i$ for each $i$. We want to pick the $f_i$ such that $$(e_1+f_1)(e_2+f_2) \cdots (e_t+f_t) = (d_1+f_1)(d_t+f_t) \cdots (d_t+f_t)$$, or $$\prod_{d_t>e_t} \frac{f_t+d_t}{f_t+e_t} = \prod_{e_t>d_t} \frac{f_t+e_t}{f_t+d_t}$$. Now let us note the following: For any nonnegative integers $(m,n,k)$ with $m>n$ and $k$ sufficiently large, we can find a positive integer $a$ such that $\frac{a+m}{a+n}=\frac{k+1}{k}$; just take $a=k(m-n)-n$ (which is positive because $k$ is large). $\frac{2^{a+k}}{2^{a+k}-1} \prod_{i=a+1}^{a+k} \frac{2^i-1}{2^i-2} = \frac{2^a}{2^a-1}$ (for example, $\frac{256}{255} \cdot \frac{255}{254} \cdot \frac{127}{126} \cdot \frac{63}{62} = \frac{32}{31}$) Combining these two observations implies that we can make both sides of the equation equal to $\frac{2^a}{2^a-1}$ for some large $a$, as desired.

It is obvious that if $k \mid n$ or $n \mid k$, that we don't have a solution. Denote by $a=(\alpha_1,\alpha_2,\alpha_3,\dots)$ the canonical factorization of the number $a$, where $\alpha_1$ is the exponent of $p_1=2$, $\alpha_2$ is the exponent of $p_2=3$, etc. . . . (we can have $0$, in fact after a point we are going to have a lot of zeroes after that) Let's denote by $n=(\alpha_1,\alpha_2,\alpha_3,\dots)$, $k=(\beta_1,\beta_2,\beta_3,\dots)$ and $k=(\gamma_1,\gamma_2,\gamma_3,\dots)$ From here we have that: $$\tau(sk)=\prod_{t=1}^{\infty} \beta_t+\gamma_t$$$$\tau(sn)=\prod_{t=1}^{\infty} \alpha_t+\gamma_t$$ We want: $$\frac{\tau(sn)}{\tau(sk)}=1$$$$\prod_{t=1}^{\infty}\frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t}=1$$ By simple algebra we easily show that we can set $\gamma_t$ to something to get that: $$\frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t} = \frac{d+e}{d}$$But we must have $\alpha_t > \beta_t$.But if $\alpha_t < \beta_t$ we will have that: $$\frac{\beta_t+\gamma_t}{\alpha_t+\gamma_t}=\frac{d}{d+e}$$ Let's divide the exponents into two sets $A,B$, where $A=\{t \mid \alpha_t > \beta_t\}$ and where $B=\{t\mid\alpha_t < \beta_t\}$.Since $d$ was undefined we can practically set $d$ for our own purposes, so now in the the realm of set $A$ we will set $d=card(A).p$, and in the realm of set $B$ we will set $d=card(B).p$, so by this we are going to have the following: $$\prod_{t=1}^{\infty} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t} = \prod_{t\in A} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t} \prod_{t \in B} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t}$$Because of the the lemma we have that: $$\prod_{t\in A} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t}=\frac{card(A)p+1}{card(A)p}\frac{card(A)p+2}{card(A)p+1}\dots \frac{card(A)p+card(A)}{card(A)p+card(A)-1}=\frac{p+1}{p}$$Similarily we have that: $$\prod_{t \in B} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t} = \frac{card(B)p}{card(B)p+1}\frac{card(B)p+1}{card(B)p+2}\dots \frac{card(B)p+card(B)-1}{card(B)p+card(B)} = \frac{p}{p+1}$$ From here we see that: $$\prod_{t=1}^{\infty} \frac{\alpha_t+\gamma_t}{\beta_t+\gamma_t} = \frac{p+1}{p}\frac{p}{p+1}=1$$ In the canonical factorisation we are going to have a lot of zeros after a while, just set $\gamma$ after that point all to be equal to $1$, just to escape the $0/0$ situation. So we have found a construction for $s$ when $n \nmid k$ and $k \nmid n$. So the answer is all pairs $(n,k)$ such that $n \nmid k$ and $k \nmid n$ . . .

I claim that all distinct positive integers $(m, n)$ for which $m \nmid n$ and $n\nmid m$ satisfy the given property. $\emph{\textbf{Proof:}}$ Let the prime factorizations of $m$ and $n$ be $$m = p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}q_1^{f_1+y_1}q_2^{f_2+y_2}\cdots q_l^{f_l+y_l}$$$$n = p_1^{e_1+x_1}p_2^{e_2+x_2}\cdots p_k^{e_k+x_k}q_1^{f_1}q_2^{f_2}\cdots q_l^{f_l}$$For positive integers $x_i$ and $y_i$ where $k, l \ge 1$. Observe that any common prime factors of $m$ and $n$ with equal exponents can be left out as they do not affect the proof. Thus we can write $sm$ and $sn$ as $$sm = p_1^{g_1-1}p_2^{g_2-1}\cdots p_k^{g_k-1}q_1^{h_1-1+y_1}q_2^{h_2-1+y_2}\cdots q_l^{h_l-1+y_l}$$$$sn = p_1^{g_1-1+x_1}p_2^{g_2-1+x_2}\cdots p_k^{g_k-1+x_k}q_1^{h_1-1}q_2^{h_2-1}\cdots q_l^{h_l-1}$$where $g_i-1 \ge e_i$ and $h_i-1 \ge f_i$. By setting $\tau(sm) = \tau(sn)$ and dividing, we obtain: \begin{align} \frac{g_1}{g_1+x_1}\cdot\frac{g_2}{g_2+x_2}\cdots\frac{g_k}{g_k+x_k} = \frac{h_1}{h_1+y_1}\cdots\frac{h_k}{h_k+y_k} \end{align} $\textbf{Definition 1}$: Define an $\emph{expand}$ as an operation in which a fraction $\frac{a}{a+1}$ is replaced by the product \begin{align*} \frac{a}{a+1} &= \frac{a+1}{a+2}\cdot \frac{a^2+2a}{a^2+2a+1}\\[2mm] &= \frac{b}{b+1}\cdot \frac{c}{c+1} \end{align*}for positive integers $a, b, c$ where $b = a+1$ and $c = a^2+2a$. Observe that $b, c > a$. $\textbf{Definition 2}$: Define an $\emph{increase}$ as an operation in which a fraction $\frac{a}{a+1}$ is replaced by the fraction $$\frac{a}{a+1} = \frac{da}{da+d} = \frac{b}{b+d}$$for positive integers $a, b$ where $b=da$. Thus if we begin with the equation $$\frac{T}{T+1} = \frac{T}{T+1} \qquad \forall \ T > \textrm{max}(e_1,\cdots, e_k, f_1, \cdots , f_l)$$we can apply the $\emph{expand}$ and $\emph{increase}$ operations on either side of the equation to obtain (1), as desired.

For OTIS! The answer is all $(m, n)$ such that $m \nmid n$ and $n \nmid m$. Clearly $m \mid n$ or $n \mid m$ don't work since $sm \mid sn$ or $sn \mid sm$ respectively, so one will always have more divisors than the other for any $s$. Now let $(e_1, e_2, \cdots)$ represent $p_1^{e_1}p_2^{e_2}\cdots$ where $p_1=2$, $p_2=3$, $\cdots$. Write $m = (m_1, m_2, \cdots)$ and $n = (n_1, n_2, \cdots)$ (all $m_i$ and $n_i$ are nonnegative integers). We wish to show that there exists nonnegative integers $s_1$, $s_2$, $\cdots$ such that $$\prod_{i=1}^{\infty} \frac{m_i+s_i+1}{n_i+s_i+1} = 1$$so then we could let $s = (s_1, s_2, \cdots)$. Acknowledge the following: Claim. For nonnegative integers $a$ and $b$ with $a>b$, it is possible to choose a nonnegative integer $t$ such that there exists a positive integer $k$ such that $\frac{k+1}{k}=\frac{a+t+1}{b+t+1}$. Proof. This is just possible by taking $t=k(a-b)-b-1$. $\square$ Now let $G$ be the set of $i$ such that $m_i > n_i$, and define $L$ similarly for $m_i < n_i$. The $i$ for $m_i = n_i$ are irrelevant, so we need only consider the $i$ that are in $G$ and $L$. It is important that $G$ and $L$ are nonempty since $m \nmid n$ and $n \nmid m$. Choose a positive integer $C$. By the claim, it is possible to create $$\prod_{i \in G} \frac{m_i+s_i+1}{n_i+s_i+1} = \frac{C|G|+1}{C|G|}\cdot\frac{C|G|+2}{C|G|+1}\cdot\cdots\cdot\frac{C|G|+|G|}{C|G|+|G|-1} = \frac{C+1}{C}$$This is only possible since $G$ is nonempty. Since $L$ is also nonempty, it is possible to make $$\prod_{i \in L} \frac{m_i+s_i+1}{n_i+s_i+1} = \frac{C}{C+1}$$by replacing every $|G|$ with $|L|$, and flipping each fraction since $m_i < n_i$. Now $\frac{C+1}{C} \cdot \frac{C}{C+1} = 1$, which finishes the problem. $\blacksquare$

Suppose $m \neq n$. The answer is $\boxed{\text{all pairs } (m,n)\text{ such that } m \nmid n,n \nmid m.}$ Obviously if $m \mid n$ or $n \mid m$, there exists no such $s$, so it suffices to construct a $s$ for all other $(m,n)$. \bigskip Note that we can simply delete all the primes in $m$ and $n$ that have equal power, so assume that $m = a_1^{p_1}\cdots a_i^{p^i}b_1^{q_1}\cdots b_j^{q_j}$ and $n = a_1^{x_1}\cdots a_i^{x^i}b_1^{y_1}\cdots b_j^{y_j}$ where $a_i,b_j$ are primes and $p_i > x_i, q_1 > y_i$. Now let $S$ be a sufficiently large integer, and let $s_k$ for $1 \leq k \leq i$ be so that $s_k = (Si + k)(a_k - b_k) + (1 - b_k)$. Define $t_k$ for $1 \leq k \leq j$ similarly, and let $s = a_1^{s_1}\cdots a_i^{s^i}b_1^{t_1}\cdots b_j^{t_j}$. Now note that $$\frac{\tau(sm)}{\tau(sn)} = \prod_{k=1}^i \frac{p_k+s_k+1}{x_k+s_k+1} \prod_{l=1}^j \frac{q_l + t_l+1}{y_l+t_l+1} = \prod_{k=1}^i \frac{iS+k+1}{iS+k} \prod_{l=1}^j \frac{jS+l}{jS+l+1} = \frac{S+1}{S} \cdot \frac{S}{S+1} = 1.$$

IndoMathXdZ wrote: Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

The answer is all pairs $(n, k)$ such that $n \nmid k$ and $k \nmid n$. The others fail since if WLOG $n \mid k$, every divisor of $sn$ divides $sk$ and $sk$ divides $sk$ as well, so there are more divisors of $sk$ than $sn$. To show that these $n, k$ work, we begin by claiming that for $N$ large enough, there exist an $a$ such that \[ \frac{x+a}{y+a} = \frac{N}{N+1},\]for all integers $x,y$ with $x<y$. The proof is simple; we can just take $a=(y-x)N-x$. Let $P$, $Q$, $R$ be sets such that for each prime $p$ dividing $n$ or $k$, it is in $P$ if $\nu_p(n) < \nu_p(k)$, in $Q$ if $\nu_p(n) = \nu_p(k)$, and in $R$ if $\nu_p(n) > \nu_p(k)$. Note that none of the the primes in $Q$ matter. For any arbitrary prime $p \in P \cup R$, it contributes a factor of \[ \frac{\nu_p (n) + \nu_p (s) + 1}{\nu_p (k) + \nu_p (s) + 1} \]to the ratio the number of divisors of $sn$ to $sk$. But by the claim, we can carefully choose $\nu_p(s)$ so that for each $p$ in $P$, this factor is equal to $\frac{N}{N+1}$ for some $N$ very large. For each $p$ in $R$, the reciprocal of the claim gives that the factor can become $\frac{N+1}{N}$ for some $N$ very large. Now to finish, if we pick some $M \gg |P|+|R|$, we can pick the values of $\nu_p(s)$ such that the ratio the number of divisors of $sn$ to $sk$ is \[ \left(\frac{M|P|-|P|}{M|P|-|P|+1} \cdot \frac{M|P|-|P|+1}{M|P|-|P|+2} \cdot \hdots \cdot \frac{M|P|-1}{M|P|} \right) \left(\frac{M|R|-|R|+1}{M|R|-|R|} \cdot \frac{M|R|-|R|+2}{M|R|-|R|+1} \cdot \hdots \cdot \frac{M|R|}{M|R|-1} \right).\]But the left product telescopes to $\frac{M-1}{M}$ and the right telescopes to $\frac{M}{M-1}$, so the product is $1$, as desired. $\blacksquare$

Answer : The answer is all pairs such that $n \nmid k$ and $k\nmid n$. Claim 1 : Given positive integers $a, b$ with $a<b$ and for any sufficiently large positive integer $A$ there exists $x$ such that $\frac{A+1}{A} = \frac{x+a}{x+b}$. Proof : $\frac{A}{A+1} = \frac{x+a}{x+b} \iff x = Ab - (A+1)a$. Since $A$ is large, $Ab - (A+1)a$ is positive. Claim 2 : Given two sequences of positive integers $\{a_i\}_{i=1}^{2^k}$ and $\{b_i\}_{i=1}^{2^k}$ such that $a_i < b_i$ and any sufficiently large $A$, there exists a sequence of positive integers $\{x_i\}_{i=1}^{2^k}$ such that for any positive integer $c$ : (i) $\prod_{i=1}^{2^k}\frac{x_i+a_i}{x_i+b_i} = \frac{2^cA}{2^cA + 1}$. (ii) $\prod_{i=1}^{2^k}\frac{x_i+a_i}{x_i+b_i} = \frac{2^cA+1}{2^cA + 2}$. Proof (i): Take $\frac{x_i+a_i}{x_i+b_i} = \frac{2^{kc}A+(i-1)}{2^{kc}A+i}$. Then \[\frac{x_1+a_1}{x_1+b_1}\cdots\frac{x_{2^k}+a_{2^k}}{x_{2^k}+b_{2^k}} = \frac{2^{kc}A}{2^{kc}A+1}\frac{2^{kc}A+1}{2^{kc}A+2}\cdots \frac{2^{kc}A+(2^k-1)}{2^{kc}A+2^k} = \frac{2^{kc}}{2^{kc}+2^k}=\frac{2^cA}{2^cA+1}.\] Proof (ii): Take $\frac{x_i+a_i}{x_i+b_i} = \frac{2^{kc}A+(2^k+i-1)}{2^{kc}A+2^k+i}$. Then \[\frac{x_1+a_1}{x_1+b_1}\cdots\frac{x_{2^k}+a_{2^k}}{x_{2^k}+b_{2^k}} = \frac{2^{kc}A+2^k}{2^{kc}A+2^k+1}\frac{2^{kc}A+2^k+1}{2^{kc}A+2^k+2}\cdots \frac{2^{kc}A+(2^{k+1}-1)}{2^{kc}A+2^{k+1}}=\frac{2^{kc}A+2^k}{2^{kc}A+2^{k+1}}=\frac{2^cA+1}{2^cA+2}.\] Claim 3 : Given two sequences of positive integers $\{a_i\}_{i=1}^{k}$ and $\{b_i\}_{i=1}^{k}$ such that $a_i < b_i$ and any sufficiently large $A$, there exists a sequence of positive integers $\{x_i\}_{i=1}^{k}$ such that : \[\prod_{i=1}^{k}\frac{x_i+a_i}{x_i+b_i} = \frac{A}{A+1}\]. Proof : Let $k = 2^{\alpha_1} + 2^{\alpha_2} + \cdots + 2^{\alpha_c}$ where $\alpha_i$ are distinct, nonnegative and $\alpha_1 < \alpha_2 < \cdots < \alpha_c$. Let $\prod_{i=1}^{2^{\alpha_1}}\frac{x_i+a_i}{x_i+b_i} = \frac{2^{c-1}A}{2^{c-1}A+1}$ $\prod_{i=\alpha_1+1}^{\alpha_2}\frac{x_i+a_i}{x_i+b_i} = \frac{2^{c-1}A+1}{2^{c-1}A+2}$ $\prod_{i=\alpha_2+1}^{\alpha_3}\frac{x_i+a_i}{x_i+b_i} = \frac{2^{c-2}A+1}{2^{c-2}A+2}$ $.$ $.$ $.$ $\prod_{i=\alpha_{k-1}+1}^{\alpha_k}\frac{x_i+a_i}{x_i+b_i} = \frac{2A+1}{2A+2}$ Hence \[\prod_{i=1}^{k}\frac{x_i+a_i}{x_i+b_i} = \frac{2^{c-1}A}{2^{c-1}A+1}\frac{2^{c-1}A+1}{2^{c-1}A+2}\frac{2^{c-2}A+1}{2^{c-2}A+2}\cdots\frac{2A+1}{2A+2} = \frac{A}{A+1}\]. The last equality can be proved easily by inducting on $c$. Main Problem Let \[n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r} p_{r+1}^{a_{r+1}}\cdots p_m^{a_m}\]\[k = p_1^{b_1}p_2^{b_2}\cdots p_r^{b_r} p_{r+1}^{b_{r+1}}\cdots p_m^{b_m}\]\[s = p_1^{x_1 - 1}p_2^{x_2 - 1} \cdots p_m^{x_m - 1}\]where $a_i >b_i$ $\forall i$ $\in \{1,2, \cdots, r\}$ and $a_j<b_j$ $\forall j \in \{r+1, r+2, \cdots, m\}$. $\tau(sm)=\tau(sn) \iff \frac{x_1+b_1}{x_1+a_1}\cdots\frac{x_r+b_r}{x_r+a_r} = \frac{x_{r+1}+a_{r+1}}{x_{r+1}+b_{r+1}}\cdots\frac{x_m+a_m}{x_m+b_m}$. However by Claim $3$, for any sufficiently large positive integer $A$, we can make the LHS and RHS both equal to $\frac{A}{A+1}$.

I claim that this is true of all $(n,k)$ such that neither $n$ nor $k$ are proper divisors of each other. Firstly, if wlog $n$ is a proper divisor of $k$, then this means that $sn$ is a proper divisor of $sk$, so $sn$ will always have strictly less divisors than $sk$, a contradiction. We now take all $n$ and $k$ such that neither are proper divisors of each other. Consider two sets, the set $A$ where $p\in A$ if $v_p(n)>v_p(k)$, and $B$ where $q \in A$ if $v_q(n)<v_q(k)$. If $n=k$, then $sn=sk$ so they will clearly have the same number of divisors. Otherwise for $n\neq k$, we clearly must have that both $A,B$ are non-empty. Next, we calculate two values, \[X = \sum_{p\in A} v_p(n)-v_p(k) \]\[Y = \sum_{q\in B} v_q(k)-v_q(n) \] We then take some arbitrarily large $M$. Additionally, we will order the primes $A$ and $B$ calling them $p_i$ with exponent differences $e_i$ for all $i\geq 1$. We achieve equal number of divisors by selecting $s$ such that \[\prod_{p\in A} \frac{v_p(ns)+1}{v_p(ks)+1} = \prod_{i=1}^{|A|} \frac{MX+\sum_{j=0}^{i} e_i}{MX+\sum_{j=0}^{i-1} e_i}= \frac{MX+X}{MX} = \frac{M+1}{M}\] By a similar process, we can do the same to get \[\prod_{q\in B} \frac{v_q(ks)+1}{v_q(ns)+1}=. \frac{MY+Y}{MY} = \frac{M+1}{M}\] Thus, the ratio of divisors will be \[\frac{d(ns)}{d(ks)} = \text{(ratio of A-primes)} \cdot \text{(ratio of B-primes)}\cdot \text{(ratio of other primes)} = \frac{M+1}{M}\cdot \frac{M}{M+1}\cdot 1 = 1\]and we are done $\blacksquare$.

We claim that any distinct integers $n$ and $k$ such that neither one divides the other will work. It is clear that if $n|k$, for instance, $(sn)|(sk)$ for any positive integer $s$, and thus we cannot have the same number of divisors for both numbers. Therefore we restrict our attention to proving there exists $s$ for $n,k$ such that neither one divides the other. Given $n$ and $k$, first omit any primes for which $n$ and $k$ have the same number of factors of that prime, as they do not affect the proof at all. Suppose that $$n=p_1^{e_1}p_2^{e_2}\ldots p_m^{e_m}q_1^{f_1}q_2^{f_2}\ldots q_n^{f_n}$$and $$k=p_1^{e_1'}p_2^{e_2'}\ldots p_m^{e_m'}q_1^{f_1'}q_2^{f_2'}\ldots q_n^{f_n'}$$where $e_i>e_i'$ for all $1\le i\le m$ and $f_i<f_i'$ for all $1\le i \le n$. Let $$s=p_1^{g_1}p_2^{g_2}\ldots p_m^{g_m}q_1^{h_1}q_2^{h_2}\ldots q_n^{h_n}.$$Our goal is to choose exponents so that the number of factors of $ks$ and $ns$ are the same; i.e. $$\prod_{i=1}^m (e_i+g_i+1) \prod_{i=1}^n (f_i+h_i+1)=\prod_{i=1}^m (e_i'+g_i+1) \prod_{i=1}^n (f_i'+h_i+1)$$Rearrange to get $$\prod_{i=1}^m \frac{e_i+g_i+1}{e_i'+g_i+1} = \prod_{i=1}^n \frac{f_i'+h_i+1}{f_i+h_i+1}.$$ Choose $g_i$ inductively for all $2\le i\le m$ so that $$e_{i-1}+g_{i-1}+1=e_i'+g_i+1\Rightarrow g_i=g_{i-1}+(e_{i-1}-e_i').$$(To prevent any of the $g_i$ from being negative, we can make $g_1$ arbitrarily large, and we will do that later.) These choices of $g_2$ through $g_m$ force the numerator of the $i-1$th term to cancel with the denominator of the $i$th term. After these cancellations are done, we are left on the left side with the numerator of the $m$th term and the denominator of the first term, or $$\frac{e_m+g_m+1}{e_1'+g_1+1}.$$Due to the inductive definition of $g_i$, $g_m=g_1+A$ for some fixed constant $A$. Let $e_m+C+1=A_1$ and $e_1'+1=A_2$, so that the left side reduces to $$\frac{g_1+A_1}{g_1+A_2}.$$Notice that $A_2$ is a positive integer. Because $\prod_{i=1}^m \frac{e_i+g_i+1}{e_i'+g_i+1}>1$, and this is a simplification of this product, $A_1>A_2$. Similarly, the right side can be expressed as $$\frac{h_1+B_1}{h_1+B_2}$$with appropriate substitutions for $h_i$ and fixed positive integer constants $B_1>B_2$. It suffices to find arbitrarily large positive integers $g_1,h_1$ so that $$\frac{g_1+A_1}{g_1+A_2}=\frac{h_1+B_1}{h_1+B_2}\Rightarrow \frac{A_1-A_2}{g_1+A_2}=\frac{B_1-B_2}{h_1+B_2}.$$However, this can be easily done by letting $N$ be an arbitrarily large positive integer and letting $$g_1+A_2=N(A_1-A_2), h_1+B_2=N(B_1-B_2).$$Then we can make $g_1,h_1$ as large as needed to prevent any of the $g_i$ or $h_i$ from becoming negative. The proof is complete. Remarks: All the proofs in this thread are long, but that is the result of writing up tons of details for a main idea that isn't difficult to spot. The choosing of $g_i,h_i$ is motivated by the "DURR WE WANT STUFF TO CANCEL" thing, if you don't know what this is then check out vEnhance's functional equation handout here and go to section 5.

bad hard to understand proof incoming The solution is all $(n, k)$ such that $n \nmid k$ and $k \nmid n$. Let the exponent set of a natural number $n$ be $(e_1, e_2, e_3, \dots)$, where $e_i=v_{p_i}(n)+1$ and $p_i$ is the $i$-th prime. For example, the exponent set of $n=2^3\cdot 3^5\cdot 5=9720$ is $(4, 6, 2, 1, 1, \dots)$. Then we want $$\prod_{i=1}^{\infty} \frac{e_i+d_i}{f_i+d_i}=1,$$where the exponent set of $n$ is $(e_1, e_2, \dots)$, the exponent set of $k$ is $(f_1, f_2, \dots)$, and the exponent set of $s$ is $(d_1+1, d_2+1, \dots)$. Note that $e_i>0$, $f_i>0$, $d_i \geq 0$ for all $i \geq 1$. Note that if $e_i=f_i$ for some positive integer $i$ then $\frac{e_i+d_i}{f_i+d_i}=1$. Claim: If $e_i > f_i$, then $\frac{e_i+d_i}{f_i+d_i}$ can take on any fraction of the form $\frac{a+1}{a}$ such that $\frac{a+1}{a}<\frac{e_i}{f_i}$. If $e_i < f_i$ then this becomes $\frac{a}{a+1}>\frac{e_i}{f_i}$. Proof: Let $e_i,f_i \equiv r \mod e_i-f_i$ where $0 \leq r < e_i-f_i$. Then we can let $d_i=c(e_i-f_i)-r$ for some number $c \geq 1$. Therefore, our product condition is now $$\prod_{i=1}^{\infty} \frac{e_i+d_i}{f_i+d_i}=1$$where $|e_i-f_i|=1$ for all $i \geq 1$. Let a fraction $r_i=\frac{e_i}{f_i}$ be top-heavy if $e_i=f_i+1$, and bottom-heavy if $e_i=f_i-1$. It's easy to show that if all $r_i$ are top-heavy ($k \mid n$), or if all $r_i$ are bottom-heavy ($n \mid k$), then the product condition fails. From this point on, we will assume that there is at least one top-heavy fraction in the product and at least one bottom-heavy fraction. Additionally, if we have two fractions $r_i$ and $r_j$ such that $r_i$ is top-heavy and $r_j$ is bottom-heavy, then these two will cancel. So, it suffices to show that: Claim: We can cancel any two top-heavy fractions $r_x$ and $r_y$ where $x,y \geq 1$ and $x\neq y$ into a single top-heavy fraction, and similarly for any two bottom-heavy fractions. The proof follows. Proof: We want to show $$\frac{f_x+d_x+1}{f_x+d_x}\cdot \frac{f_y+d_y+1}{f_y+d_y}=\frac{a+1}{a}$$for some positive integer $a$. Choose $d_x, d_y$ such that $f_x+d_x=f_y+d_y+1=\text{some odd number}.$ Then $f_x+d_x+1=f_y+d_y+2=\text{some even number}$, so $\frac{f_x+d_x+1}{f_y+d_y}=\frac{a+1}{a}$ for some positive integer $a$. $\blacksquare$

as an example for exponent sets $n=(4, 6, 2, 1, 1, \dots)$ and $k=(3, 7, 11, 1, 2, 1, 1, \dots)$ the fractions are $\frac{4+d_1}{3+d_1}$, $\frac{6+d_2}{7+d_2}$, $\frac{1+d_3}{2+d_3}$, $\frac{1+d_4}{2+d_4}$ then we have $d_1=0$, $d_2=0$, $d_3=13$, $d_4=14$ which corresponds to $0, 0, 124, 14$ so $n=2^3\cdot 3^5\cdot 5^1\cdot 7^0\cdot 11^0$ $k=2^2\cdot 3^6\cdot 5^{10}\cdot 7^0\cdot 11^1$ $s=2^0\cdot 3^0\cdot 5^{124}\cdot 7^0\cdot 11^{14}$

Hard problem for N1

Answer : All pairs of $(m,n)$ satisfies for $m \nmid n$ and $n \nmid m$ It can be seen that for all the pairs in which $m | n $ or $n | m$ when any number is multiplied the greater of $m,n$ in the above case would have more divisors and the divisors of the smaller number would be a subset of the set of divisors of the larger one Now take the pairs of $(m,n)$ for which $m \nmid n$ and $n \nmid m$ Now if the pairs are coprime then any prime different from the prime divisors of both $m,n$ would work ( there are infinte primes) , In $m=n$ then any numbr would work Now lets take \[m = p_1^{a_1}p_2^{a_2} \dots p_e^{a_e} \]\[n = p_1^{b_1}p_2^{b_2} \dots p_e^{b_e}\]\[s = p_1^{c_1}p_2^{c_2} \dots p_e^{c_e}\]From the number of divisors of $ms,ns$ we have \[ (a_1+c_1+1)(a_2+c_2+1) \dots (a_e+c_e+1) = (b_1+c_1+1)(b_2+c_2+1) \dots (b_e+c_e+1) \]\[ \frac{(b_1+c_1+1)(b_2+c_2+1) \dots (b_e+c_e+1)}{(a_1+c_1+1)(a_2+c_2+1) \dots (a_e+c_e+1)} = 1 \] Claim : For some real $x \ge a+1$ there exist a $c$ such that \[ \frac{b+c+1}{a+c+1} = \frac{x+1}{x} \] $Proof -$ \[ \frac{b+c+1}{a+c+1} = \frac{1+x}{x}\] \[ bx+cx+1 = ax+cx+1+a+c+1 \]\[ c= x(b-a) -(a+1)\] Let $Y$ be a large number greater than all $a_i,b_i$ WLOG lets say $i$ from $1$ to $f$ , $b_i>a_i$ and from $f+1$ to $e$ , $b_i<a_i$ Case 1 : $b_i>a_i$ From the above claim we can set $c_i$ in such a way such that \[ \frac{b_i+c_i+1}{a_i+c_i+1} = \frac{fY+i}{fY+i-1} \] Case 2 : $b_i < a_i$ From the above claim we can set $c_{f+i}$ in such a way that \[ \frac{a_{f+i}+c_{f+i}+1}{b_{f+i}+c_{f+i}+1} = \frac{(e-f)Y+i}{(e-f)Y+i-1} \] Now we can say \[\prod_{i=1}^{e} \frac{b_i+c_i+1}{a_i+c_i+1} = \prod_{i=1}^{f}\frac{b_i+c_i+1}{a_i+c_i+1} \enspace \prod_{i=1}^{e-f} \frac{b_{f+i}+c_{f+i}+1}{a_{f+i}+c_{f+i}+1} \]\[ = \prod_{i=1}^{f} \frac{fY+i}{fY+i-1} \enspace \prod_{i=1}^{e-f} \frac{(e-f)Y+i-1}{(e-f)Y+i} \]\[=1\]

We claim the answer is $n$ does not divide $k$ or vice versa. Clearly, if $n$ divides $k$ then $\tau(sn)>\tau(sk)$. So, assume that does not hold. Let us construct such an $s$. Let $n=p_1^{\alpha_1}p_2^{\alpha_2} \cdots p_m^{\alpha_m}$ and $k=p_1^{\beta_1}p_2^{\beta_2} \cdots p_m^{\beta_m}$ and $s=p_1^{\gamma_1}p_2^{\gamma_2} \cdots p_m^{\gamma_m}$. We want to show that there exists such a choice of $\gamma_i$ satisfying. \[ \frac{\alpha_1+\gamma_1+1}{\beta_1+\gamma_1+1} \cdot \frac{\alpha_2+\gamma_2+1}{\beta_2+\gamma_2+1} \cdots \frac{\alpha_m+\gamma_m+1}{\beta_m+\gamma_m+1}=1\]since $n$ and $k$ do not divide one another, then there exists $x$ and $y$ where $a_x>b_x$ and $a_y<b_y$. Thus, assume wlog that $a_i<b_i$ for $1 \leq i < \ell$ and $a_i>b_i$ for $\ell \leq i \leq m$ (if $a_i=b_i$ then it is irrelevant since it cancels in our fraction). By 2004 Putnam A1 we see that for every sufficiently large positive integer $N$, the equation \[ \frac{\alpha_i+\gamma_i+1}{\beta_i+\gamma_i+1} = \frac{N}{N+1}\]has a solution for $i < \ell$ and the other way around for $i \geq \ell$. Thus, we can get our massive product from before to be \[ \frac{M}{M+(m-\ell)} \cdot \frac{M+(m-\ell)}{M+2(m-\ell)} \cdots \frac{M+(\ell-1)(m-\ell)}{M+\ell(m-\ell)} \cdot \frac{M+\ell(m-\ell)}{M+\ell(m-\ell-1)} \cdots \frac{M+\ell}{M} =1\]where $M$ is a sufficiently large positive integer divisible by both $\ell$ and $m-\ell$. $\blacksquare$

:skull: $(n,n)$ is a solution, so let $n=\neq k$. Obviously, there are no other solutions when $n\mid k$ or $k\mid n$ because then $sn \mid sk$ or $sk\mid sn.$ $~$ Let $(p_1,p_2,\dots,p_j)$ be the list of all primes, in order from smallest to largest, such that $\nu_{p_i}{n}\neq \nu_{p_i}{k}.$ Note that if $(n,k)$ then is a solution, we can multiply by any $m$ that is coprime to both $n,k$ and $\tau(smn)=\tau(smk)$ will still hold. $~$ We just need to show that as long as there exists both $\nu_{p_i}{n}>\nu_{p_i}{k}$ and $\nu_{p_i}{n}<\nu_{p_i}{k}$ then there exists an $s.$ $~$ Let $n=\prod_{i=1}^{j}{p_1^{n_i}}$ and $k=\prod_{i=1}^{j}{p_1^{k_i}}$ then $\tau(n)=\prod_{i=1}^{j}{1+n_i}$ and $\tau(k)=\prod_{i=1}^{j}{1+k_i}.$ Now, we can add an integer $r_i$ to $(1+n_i)$ and $(1+k_i)$ so that the products of $(1+n_i+r_i)$ and $(1+k_i+r_i)$ are equal. $~$ We already assumed $n_i\neq k_i$, so when $n_i>k_i$ (and analagously otherwise) solving the equation $\frac{1+n_i+r_i}{1+k_i+r_i}=\frac{M_i+1}{M_i}$ we see there is an integer $r_i$ that satisfies. Thus, let there be $a$ numbers $n_i>k_i$ and $b$ numbers $n_i<k_i$. $~$ We choose the following $M_i$: \[\prod_{i=1}^{j}\frac{1+n_i+r_i}{1+k_i+r_i}=\frac{aN+1}{aN}\cdot \frac{aN+2}{aN+1}\cdot \frac{aN+a}{aN+a-1}\cdot \frac{bN}{bN+1}\cdot \frac{bN+1}{bN+2}\cdot \frac{bN+b-1}{bN+b}\]and we win.

We claim that it holds for all $n$ and $k$ that do not divide each other Let $n=p_1^{a_1}p_2^{a_2}....p_k^{a_k}$ $k=p_1^{b_1}p_2^{b_2}....p_k^{b_k}$ $S=p_1^{x_1}p_2^{x_2}....p_k^{x_k}$ Claim : If either $n|k$ or $k|n$ it fails Proof: If $n|k$, then $a_k \leq b_k$ for all $k$ However, since $n,k$ are distinct, it implies that $ \prod (a_k +x_k+1) < \prod (b_k+x_k+1)$ for all $S$ Claim: If $n$ and $k$ do not divide each other, we win Proof: Let $F_k$ denote $\frac{a_k+x_k+1}{b_k+x_k+1}$ For the sake of brevity, suppose $F_k>1$ for $i$ values of $k$, $F_k<1$ for $j$ values of $k$ Notice that $F_k=1$ is irrelevant as it doesn't affect the product of the number of divisors Then, choose a very larger number $Y$ such that we can express $F_k$ as $\frac{Y}{Y+1}$ or $\frac{Y+1}{Y}$ for all $k$ For the $i$ values of $k$ where $F_k>1$, choose values of $x$ such that the $i$ values become $$\frac{Yi+1}{Yi},\frac{Yi+2}{Yi+1},....\frac{Yi+i}{Yi+i-1}$$ Similarly, for the $j$ values of $k$ where $F_k<1$, choose values of $x$ such that the $j$ values become $$\frac{Yj}{Yj+1},\frac{Yj+1}{Yj+2},....\frac{Yj+j-1}{Yj+j}$$ Note that the two products simply reduce to $$(\frac{Y+1}{Y})(\frac{Y}{Y+1}) =1$$ Hence, we are done

For each $(n,k)$ we can remove all the primes in the prime factorization that have the same exponent in $n$ and $k$. This clearly does not affect anything(any modification by multiplying that prime is mirrored exactly in the other number). Call this new pair $(a,b)$. If $a=b=1$, this clearly works. If $a|b$ or $b|a$ but not $a=b=1$, this is clearly impossible. We claim that all other cases are possible. First of all, we can see that there must be at least one $p$ such that $\nu_p(a)<\nu_p(b)$ and at least one $q$ such that $\nu_q(a)>\nu_q(b)$(this is implied by the previous assumptions that $a\nmid b$ and $b\nmid a$). For the cases where $\nu_p(a)<\nu_p(b)$, we can try to make the $p$ part in the ratio of the number of divisors formula between $a$ and $b$ equal something very close to but below $1$ by selecting a large $e$ and multiplying $p^e$ to both $a$ and $b$. Specifically, there exists a positive integer $x$ such that for all $y\ge x$, the ratio can be $\frac{y}{y+1}$. This is because if we have $$e=y(\nu_p(b)-\nu_p(a))-\nu_p(a)$$for a large enough $y$, this works. A similar argument can be applied to when $\nu_q(a)>\nu_q(b)$, and we can get ratios of the form $\frac{y+1}{y}$. Finally, we can force the ratio to be $1$ by selecting appropriate $y$s such that the product of the $\frac{y}{y+1}$s cancel out with the product of the $\frac{y+1}{y}$s. If there are $s$ $p$s such that $\nu_p(a)<\nu_p(b)$ and $t$ $q$s such that $\nu_q(a)>\nu_q(b)$, there exists a large enough $z$ such that the fractions would become $$\left(\frac{sz}{sz+1}\cdot\frac{sz+1}{sz+2}\cdot\dotso\cdot\frac{sz+s-1}{sz+s}\right)\cdot\left(\frac{tz+t}{tz+t-1}\cdot\frac{tz+t-1}{tz+t-2}\cdot\dotso\cdot\frac{tz+1}{tz}\right)=1.$$Therefore, this works if and only if $a=b=1$ or ($a\nmid b$ and $b\nmid a$), I'm using parentheses because the logic expression order isn't very clear. This translates to $n=k$ or ($n\nmid k$ and $k\nmid n$). QED.

We claim all $(n,k)$ that satisfy none or both of $n\mid k$ and $k\mid n$ work. Clearly, if $n\mid k$ or $k\mid n$ but $n\neq k$ implies $\tau(sn)\neq\tau(sk),$ and if $n=k,$ then $sn=sk.$ Suppose we have the prime factorizations $n=\prod_{i=1}^{\infty}p_i^{\alpha_i},$ $m=\prod_{i=1}^{\infty}p_i^{\beta_i},$ and $s=\prod_{i=1}^{\infty}p_i^{\gamma_i-1},$ where for at least one $i$ we have $\alpha_i>\beta_i$ and for at least one $i$ we have $\alpha_i<\beta_i.$ Define $\delta_i$ as $\alpha_i-\beta_i$ and assume WLOG that for $1\le i\le k-1,$ we have $\delta_i>0,$ for $k\le i\le \ell-1$ we have $\delta_i<0,$ and for $i\ge\ell,$ we have $\delta_i=0.$ Finally, let $K_1=\delta_1+\dots+\delta_{k-1}$ and $K_2=|\delta_k|+\dots+|\delta_{\ell-1}|.$ Choose $\gamma_i=A-\beta_i+\delta_1+\dots+\delta_{i-1}$ for $1\le i\le k-1$ where $A$ is a sufficiently large multiple of $K_1.$ Then, $$\prod_{i=1}^{k-1}\frac{\alpha_i+\gamma_i}{\beta_i+\gamma_i}=\prod_{i=1}^{k-1}\frac{A+\delta_1+\dots+\delta_i}{A+\delta_1+\dots+\delta_{i-1}}=\frac{A+K_1}{A}.$$Choose $\gamma_i=AK_2/K_1-\alpha_i+|\delta_k|+\dots+|\delta_{i-1}|$ for $k\le i\le \ell-1$ for $$\prod_{i=k}^{\ell-1}\frac{\alpha_i+\gamma_i}{\beta_i+\gamma_i}=\prod_{i=k}^{\ell-1}\frac{AK_2/K_1+|\delta_k|+\dots+|\delta_{i-1}|}{AK_2/K_1+|\delta_k|+\dots+|\delta_i|}=\frac{AK_2/K_1}{AK_2/K_1+K_2}.$$Hence, $$\prod_{i=1}^{\infty}\frac{\alpha_i+\gamma_i}{\beta_i+\gamma_i}=\frac{A+K_1}{A}\cdot\frac{AK_2}{AK_2+K_1K_2}=1,$$as desired. $\square$

We claim it is possible when either $m=n$ or $m\nmid n$ and $ n \nmid m.$ If $m\mid n$ or $n\mid m$ and $m\neq n$, it is obvious why it doesn't work. Let $m = \prod_{i=1}^k p_i^{a_i}, n = \prod_{i=1}^k p_i^{b_i}, s = \prod_{i=1}^k p_i^{c_i}.$ We want to show that it is possible to choose $c_i$ such that $$\prod_{i=1}^k \frac{a_i + c_i+1}{b_i + c_i + 1} = 1.$$Without loss of generality, say that $b_1> a_1, b_2>a_2, \cdots, b_j>a_j$ and $b_{j+1} < a_{j+1}, b_{j+2}<a_{j+2}, \cdots, b_k < a_k$ for some $j$. Then choose $c_i$ such that there exists some arbitrarily large $X$ such that for $1\leq i \leq j$, $\frac{a_i+c_i+1}{b_i+c_i+1} = \frac{jX+i-1}{jX+i}$ and for $j+1 \leq i \leq k$, $\frac{a_i+c_i+1}{b_i+c_i+1} = \frac{(k-j)X + i - j}{(k-j)X + i-j-1}.$ Then $$\prod_{i=1}^k \frac{a_i+c_i+1}{b_i+c_i+1} = \prod_{i=1}^j \frac{a_i+c_i+1}{b_i+c_i+1}\prod_{i=j+1}^k \frac{a_i+c_i+1}{b_i+c_i+1} = \frac{X}{X+1} \cdot \frac{X+1}{X} =1.$$

The answer is all $m, n$ such that $m \nmid n$ and $n \nmid m$. Obviously, these cases fail. Now, it suffices to show that we can choose nonnegative integers $k_1, k_2, \dots, k_n$ such that $$\prod_{i=1}^n \frac{k_i}{k_i+e_i} = 1$$for any integers $e_1, e_2, \dots, e_n$ that are not all the same sign. This is achievable by choosing some constants $\alpha_1, \alpha_2, \dots, \alpha_n$ such that the equations $$\alpha_i(k_i+e_i) = \alpha_{i+1}k_{i+1}$$are true for all $1 \leq i \leq n$, where indices are cyclic. By summing all the equations, it is sufficient for these $\alpha_i$ to satisfy $$\sum_{i=1}^n \alpha_i e_i = 0.$$To construct this, without loss of generality let $\sum_{i=1}^n e_i$ be positive. (If it is zero, then take all the $\alpha_i$ to be equal), and furthermore suppose $e_n$ is negative. Then, set $\alpha_1 = \alpha_2 = \cdots = \alpha_{n-1} = \alpha$, such that $$\alpha_n = \frac{\alpha\sum_{i=1}^{n-1} e_i}{-e_n}.$$By choosing $e_n \mid \alpha$, we can guarantee that all the $\alpha_i$ are positive integers. Observe that this means that all the $k_i$ for $1 \leq i \leq n-1$ can be expressed in the form $k_1+c$ for some integer $c$, so by choosing $k_1$ sufficiently large, they can all be positive integers. Finally, the equation $$\alpha(k_{n-1} + e_{n-1}) = \alpha_n k_n$$will yield a positive integer solution for $k_n$ as long as $$\alpha_n \mid k_{n-1} + e_{n-1} = k_1+c+e_{n-1}.$$Thus, picking $k_1 \equiv -c \pmod {\alpha_n}$ suffices. As a result, such $k_i$ must exist, so we are done.

Let \begin{align*} m = p_1^{a_1}\dots p_k^{a_k}\\ n = p_1^{b_1}\dots p_k^{b^k}\\ s = p_1^{c_1}\dots p_k^{c^k} \end{align*}If $sm$ and $sn$ have an equal number of divisors note that, $$(a_1+c_1+1)\dots (a_k+c_k+1) = (b_1+c_1+1)\dots(b_k+c_k+1)$$which implies $$S=\frac{a_1+c_1+1}{b_1+c_1+1}\dots \frac{a_k+b_k+1}{a_k+c_k+1}=1$$Then, notice that $b_i+c_i+1 = a_i+c_i+1-(a_i-b_i)$. Then, letting $|a_i-b_i|=d_i$ and $a_i+c_i+1=e_id_i$. We have that $$S = \prod_{i=1}^k \frac{d_ie_i}{d_ie_i \pm d_i}=\prod_{i=1}^k \frac{e_i}{e_i \pm 1}$$where the plus or minus is decided based on which of $a_i$ and $b_i$ is greater. If all of these are pluses, then the product will be strictly less than 1. If all of them are minuses, then the product will be strictly greater than 1. \begin{claim*} If one of them is $+$ and the other is $-$, then a valid solution exists. \end{claim*} Notice that then, we have $$S = \frac{e_1}{e_1+1}\frac{e_2}{e_2-1}\dots \frac{e_k}{e_k \pm 1}$$Now, we have two possible cases, \textit{Case 1 :} $$\frac{e_1}{e_1+1}\frac{e_2}{e_2-1} = \frac{p}{q}>1$$Then, $\frac{q}{p}< 1$, then we take all $e_i \pm 1$ to have $+$. Next say that $\frac{q}{p}$ can be written as the product of some $k-2$ rational numbers taking the numerator and denominator as required. The other case is similar. Thus, if $v_p(m)>v_p(n)$ for some prime $p$ in $m,n$ and $v_q(m)<v_q(n)$ for another prime $q \neq p$, there exists some $s$ which satisfies the required conclusion.

Assume $m \neq n$ from now on because otherwise you can take any $s$. We claim that the answer is all pairs $(m,n)$ such that $m \nmid n$ and $n \nmid m$. Clearly, if $m \mid n$ or $n \mid m$, there exist no value of $s$. We will prove that otherwise, there must exist a positive integer $s$. Claim: If $\alpha > \beta \ge 0$ are integers then for any $M \ge \beta+1$ there exists a positive integer $x$ such that \[ \frac{\alpha+x+1}{\beta+x+1} = \frac{M+1}{M}. \] Proof: Setting $x = M\alpha-(M+1)\beta+1$ works. $\square$ Now, suppose for primes we write \begin{align*} m &= p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k} q_1^{\gamma_1} \dots q_\ell^{\gamma_\ell} \\ n &= p_1^{\beta_1} p_2^{\beta_2} \dots p_k^{\beta_k} q_1^{\delta_1} \dots q_\ell^{\delta_\ell} \end{align*} where $\alpha_i > \beta_i$ and $\gamma_j < \delta_j$. We have disregarded any $p$ for which $\nu_p(m) = \nu_p(n)$, since these do not affect the proof in any way. Let $T$ be a large integer exceeding $\max(\alpha_i, \beta_j)$. We can pick $x_i$ for $i=1,\dots,k$ such that \[ \prod_i \frac{x_i + \alpha_i + 1}{x_i + \beta_i + 1} = \frac{kT+1}{kT} \cdot \frac{kT+2}{kT+1} \cdot \dots \cdot \frac{kT+k}{kT+(k-1)} = \frac{kT+k}{kT} = \frac{T+1}{T}. \] Similarly we can pick $y_i$ for $i=1,\dots,\ell$ such that \[ \prod_j \frac{y_j + \delta_j + 1}{y_j + \gamma_j + 1} = \frac{\ell T+1}{\ell T} \cdot \frac{\ell T+2}{\ell T+1} \cdot \dots \cdot \frac{\ell T+\ell }{\ell T+(\ell -1)} = \frac{\ell T+\ell }{\ell T} = \frac{T+1}{T}. \] Then if we let $s = \prod_i p_i^{x_i} \prod_i q_i^{y_i}$, we have \[ \frac{d(sm)}{d(sn)} = \prod_i \frac{x_i + \alpha_i + 1}{x_i + \beta_i + 1} \prod_j \frac{y_j + \gamma_j + 1}{y_j + \delta_j + 1} = \frac{T+1}{T} \cdot \frac{T}{T+1} = 1 \] as desired. $\square$

For mindset, replace $k$ by $m$. The answer is all $m, n$ with $m \nmid n$, $n \nmid m$. Note that the only ``relevant" primes $p$ are the ones that divide at least one of $m, n$; otherwise they contribute exactly the same multiple to both $d(sn)$ and $d(sm)$, so it suffices to have $s$ divisible by only those primes. Let's assume there are $k$ primes dividing at least one of $m, n$. Let $m = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, $n = p_1^{\beta_1}p_2^{\beta_2} \cdots p_k^{\beta_k}$. Then, it suffices to exist a solution $(x_1, x_2, \cdots, x_k)$ to the equation $$\prod_{i = 1}^{k} \left(\frac{\alpha_i + x_i}{\beta_i + x_i}\right) = 1.$$Note that $\alpha_i = \beta_i$ doesn't matter, since they don't change the product anyway, so let's split all of $i \in \{1, 2, \cdots, k \}$ with $\alpha_i \ne \beta_i$ into two groups: $P$, with $\alpha_i > \beta_i$, and $N$ with $\alpha_i < \beta_i$. Let $P = \{p_1, p_2, \cdots, p_{|P|} \}$ and $N = \{n_1, n_2, \cdots, n_{|N|}\}$. So, our product transforms into $$\prod_{i \in P} \left(\frac{\alpha_i + x_i}{\beta_i + x_i}\right) \cdot \prod_{i \in N} \left(\frac{\alpha_i + x_i}{\beta_i + x_i}\right) = 1.$$Now, this transforms to $$\prod_{i \in P} \left(\frac{\alpha_i + x_i}{\beta_i + x_i}\right) = \prod_{i \in N} \left(\frac{\alpha_i + x_i}{\beta_i + x_i}\right),$$now we aim to make this ``telescope", here's how: First, pick some $x_{p_1}$. Then, we will pick inductively, in order for any $2 \le i \le |P|$, $x_{p_i}$ such that $\alpha_{p_{i - 1}} + x_{p_{i - 1}} = \beta_{p_i} + x_{p_i}$. Then, our LHS becomes $\left(\frac{V + x_{p_1}}{\beta_{p_1} + x_{p_1}} \right)$, where $V = \alpha_{p_1} + (\alpha_{p_2} - \beta_{p_2}) + \cdots + \alpha_{p_{|P|}}$. Similarly picking $y_i$ for the RHS such that it telescopes as well, we have $\left(\frac{W + x_{n_1}}{\alpha_{n_1} + x_{n_1}}\right)$, where $W = \beta_{n_1} + (\beta_{n_2} - \alpha_{n_2}) + \cdots + \beta_{n_{|N|}}$ . So, it suffices for $\frac{V - \beta_{p_1}}{\beta_{p_1} + x_{p_1}} = \frac{W - \alpha_{n_1}}{\alpha_{n_1} + x_{n_1}}$, it is obvious that there exist infinitely many solutions $(x_{p_1}, x_{n_1})$ to this; just pick any one large enough for every $x_i$ to be positive, and we're done.

If $m=n$, any $s$ works so from now on assume WLOG that $n>m.$ I claim the answer is all pairs $(m,n)$ for which $m\nmid n$ and $n\nmid m$. Firstly, we will show that there is no solution for $s$ if $m\mid n$. Indeed, if $m\mid n$, then $sm\mid sn$. This means that $\tau(sm)<\tau(sn)$. From now on assume $m\nmid n$. Let $p_1,p_2,\dots p_k$ be all prime divisors of $mn$. Hence, $$n=\prod_{i=1}^kp_i^{\alpha_i} \hspace{3mm}\text{and}\hspace{3mm} m=\prod_{i=1}^kp_i^{\beta_i}.$$We will search $s$ to be in the form $$s=\prod_{i=1}^kp_i^{\gamma_i},$$where we will choose $\gamma_i$ to satisfy $\tau(sm)=\tau(sn).$ Now we have $$\frac{\tau(sn)}{\tau(sm)}=\prod_{i=1}^k\frac{\alpha_i+\gamma_i+1}{\beta_i+\gamma_i+1}.$$Notice that if $\alpha_i=\beta_i$ for some $i$, then the "$i$-th" fraction will be equal to $1$ and will not change the product, so assume that $\alpha_i\neq\beta_i$ for all $i$. Let us prove the following claim. Claim: Let $\alpha>\beta$. Then, if $M$ is any integer such that $M\ge\beta+1$, there exists an integer $\gamma$, for which $$\frac{1+\alpha+\gamma}{1+\beta+\gamma}=\frac{M+1}{M}.$$ Proof: Indeed, the condition above is equivalent to $$\gamma = M(\alpha-\beta)-(\beta+1),$$so we are done. WLOG let $\alpha_i>\beta_i$ for $i=1,2\dots,r$ for some natural $r$ and let $\alpha_j<\beta_j$ for $j=r+1,r+2,\dots,k$. Now pick a sufficiently large integer $T$. Then from the claim we have $$\prod_{i=1}^r\frac{1+\alpha_i+\gamma_i}{1+\beta_i+\gamma_i}=\frac{rT+1}{rT}\cdot\frac{rT+2}{rT+1}\dots\frac{rT+r}{rT+r-1}=\frac{T+1}{T}.$$Similarly, we obtain $$\prod_{i=r+1}^k\frac{1+\beta_i+\gamma_i}{1+\alpha_i+\gamma_i}=\frac{(k-r)T+1}{(k-r)T}\cdot\frac{(k-r)T+2}{(k-r)T+1}\dots\frac{(k-r)T+(k-r)}{(k-r)T+k-r-1}=\frac{T+1}{T}.$$ Multiplying these two gives that $$\frac{\tau(sn)}{\tau(sm)}=\frac{T+1}{T}\cdot\frac{T}{T+1}=1,$$so we are done.

Let $m$, $n$, and $p$ be in the form \[ m = \prod_i^k p_i^{a_i}\]\[n = \prod_i^k p_i^{b_i}\]\[s = \prod_i^k p_i^{c_i}\]with $a_i$ or $b_i$ possibly equaling $0$ but not both. Notice that if $m | n$ or vice versa with $m \neq n$ then obviously $sm$ and $sn$ can't have the same number of divisors. If $m = n$ then $s = 1434$ works. Now if $a_i = b_i$ for any $i$ the we can effectively remove it from the sequence for obvious reasons. We can also let $\alpha_i$ be $a_i$ but ordered least to greatest with $\beta_i$ being defined similarly for $b_i$. Let $p$ be an integer so that $\alpha_p < \beta_p$ but $\alpha_{p+1} > b_{p+1}$. So then we can set $c_i$ so that for some sufficiently large integer $\mathcal{Q}$ we have \[\prod_{i=1}^{p} \frac{\alpha_i + c_i + 1}{\beta_i + c_i + 1} = \prod_{i=1}^{p} \frac{p\mathcal{Q} + i - 1}{p\mathcal{Q} + i} = \frac{\mathcal{Q}}{\mathcal{Q} + 1}\]and \[\prod_{i={p+1}}^{k} \frac{\alpha_i + c_i + 1}{\beta_i + c_i + 1} = \prod_{i=p+1}^{k} \frac{(k-p)\mathcal{Q} + i - p}{(k-p)\mathcal{Q} + i - p - 1} = \frac{\mathcal{Q} + 1}{\mathcal{Q}}\]so we get that $\prod_i \frac{a_i + c_i + 1}{b_i + c_i + 1} = 1$ as desired.

I claim the answer is only pairs $(n,k)$ such that $n$ does not divide $k$ and $k$ does not divide $n$. Proof claimed pairs work: Let $n = p_1^{a_1} \cdots p_y^{a_y} q_1^{c_1} \cdots q_z^{c_z}, k = p_1^{b_1} \cdots p_y^{q_y} q_1^{d_1} \cdots q_z^{d_z}$, let $s = p_1^{e_1} \cdots p_y^{e_y} q_1 \cdots q_z^{f_z}$, where $a_i \ge b_i$ and $c_i \le d_i$. Then we desire to prove $\prod \frac{e_i + a_i + 1}{e_i + b_i + 1} = \prod \frac{f_i + d_i + 1}{f_i + c_i + 1}$. Choosing such $e_i, f_i$ is easy. We claim that we can achieve each side of the product being $\frac{x + 1}{x}$ for sufficiently large $x$. Each individual term in the product can hit $\frac{m + 1}{m}$ for all sufficiently large integers $m$, so we can chain and write the left side as $\frac{m + y}{m}$, so we can achieve $\frac{x + 1}{x}$ for all sufficiently large $x$, finishing. Proof nothing else works: Clearly all divisors of one of the numbers are all divisors of the other, and since one number is larger it has more divisors.

Needed to use hint The answer is all $m, n$ such that one does not divide the other (except for $m=n$ ofc). Let $v_{p_i}(m)=e_i, v_{p_i}(n)=f_i, v_{p_i}(s)=g_i$. Then we want $$\prod \frac{e_i+g_i+1}{f_i + g_i+1} = 1$$Now when all $e_i \leq f_i$ or vice versa without them all being equal it is obviously not possible because every term is $\leq 1$ so the product cannot possibly be $1$ without them all being equal. Call a factor of that product down if $e_i \leq f_i$ and up if $e_i > f_i$. Claim: For up factors we can choose a sufficient $g$ such that it is in the form $\frac{M+1}{M}$, and simiarly for down factors. Proof: You can just find a solution of the linear equation that follows. If there are $x$ up factors and $y$ down factors then we can group all of them together to telescope the product into $$\frac{M_1}{M_1+x} \frac{M_2+y}{M_2}$$for any integers $M_1, M_2$ by our claim. Setting $M_1 = y, M_2=x$ clearly works so we are done.

Simplest solution By that I mean non rigorous Infinite d(n) = d(k) n = p1^e1 •p2^e2 • ....... • pz^ez k = g1^f1 •g2^f2 • ....... • gz^fz Here not all gi^fi = pi^ei If s does not divide n,k Then d(sn) = d(sk) If s | k or n or both Then if s = x1^w1 • ........• xh^wh Some of xi will match with pi These same xi will match with gi Therefore there will be infinite such numbers

I was starting to run of variable names Replace $(n,k)$ by $(a,b)$ and $s$ by $t$. We claim that there exist a positive integer $t$ satisfying $\tau(ta)=\tau(tb)$ if and only if $a \nmid b$ and $b\nmid a$. If $a\mid b$, since $a\ne b$, $ta$ is a proper divisor of $tb$. Hence, $\tau(ta) < \tau(tb)$, and no such $t$ exists. By symmetry, the case where $b\mid a$ is analogous, so we now show that if $a \nmid b$ and $b\nmid a$, then there is some $t$ such that $\tau(ta)=\tau(tb)$. We first introduce some variables. Let $a=p_1^{e_1}\cdots p_r^{e_r}o_1^{c_1}\cdots o_u^{c_u}$ and $b=q_1^{f_1}\cdots q_s^{f_s}o_1^{d_1}\cdots o_u^{d_u}$, where $r$, $s$, and $u$ are positive integers, $e_1,\dots, e_r, f_1, \dots, f_s, c_1, \dots, c_u, d_1, \dots, d_u$ are positive integers, and $p_1,\dots, p_r, q_1, \dots, q_s, o_1, \dots, o_u$ are distinct primes. Then, let $t = p_1^{g_1-1}\cdots p_r^{g_r-1}q_1^{h_1-1}\cdots q_s^{h_s-1}o_1^{k_1-1}\cdots o_u^{k_u-1}$, where $g_1,\dots, g_r, h_1, \dots, h_s, k_1, \dots, k_u$ are positive integers. In order to have $\tau(ta)=\tau(tb)$, we must have \begin{align*} ((e_1+g_1)\cdots (e_r+g_r)) \cdot (h_1 \cdots h_s) \cdot ((c_1+k_1)\cdots (c_u+k_u)) &= \\ ((f_1+h_1)\cdots (f_s+h_s)) \cdot (g_1 \cdots g_r)\cdot ((d_1+k_1)\cdots (d_u+k_u)).\end{align*}Let $A$ be the set of ordered pairs $(c_i,d_i)$ where $c_i > d_i$, and let $B$ be the set of ordered pairs $(c_i,d_i)$ where $c_i<d_i$. If $c_i = d_i$, the terms $c_i+k_i$ and $d_i+k_i$ cancel in the above equation, and we can ignore the values of $c_i$ and $d_i$. Now, for all $(c_i,d_i)\in A$, let $k_i = (c_i-d_i)v_i - d_i$, where every $v_i$ is a positive integer, and for all $(c_i,d_i)\in B$, let $k_i = (d_i-c_i)w_i - c_i$, where every $w_i$ is a positive integer. Also, for every valid $i$, let $g_i = x_ie_i$ for some positive integer $x_i$, and for every valid $i$, let $h_i = y_ih_i$. The above equation can now be rearranged as \[\left(\prod_{i=1}^r \frac{x_i+1}{x_i} \right)\left(\prod_{(c_i,d_i)\in A} \frac{v_i+1}{v_i}\right) = \left( \prod_{i=1}^s \frac{y_i+1}{y_i}\right)\left(\prod_{(c_i,d_i)\in B} \frac{w_i+1}{w_i}\right).\]Call a positive rational number that can be written in the form $\tfrac{n+1}{n}$ for a positive integer $n$ tight. To finish the problem, note that it suffices to prove that for any two positive integers $N_1$ and $N_2$, we can find $N_1$ tight numbers that have some product $P$, and $N_2$ tight numbers that have the same product $P$. This is because we can let each $x_i$, $y_i$, $v_i$, and $w_i$ be any positive integer. To do this, we claim that any tight number can be written as the product of $m$ tight numbers, where $m$ is any positive integer. The claim is trivial when $m=1$, and for any $m>1$, consider \[\frac{n+1}{n} = \prod_{i=0}^{m-1} \frac{mn+i+1}{mn+i},\]which proves the claim. Now, pick some arbitrary tight number, and express it as a product of $r+|A|$ tight numbers and as a product of $s+|B|$ tight numbers. $\blacksquare$

If $n\mid k$ or $k\mid n$, clearly $s$ cannot exist. Conversely, we show that if $n\nmid k$ and $k\nmid n$, then it is possible to construct such a value of $s$. Let $a_1,a_2,\dots,a_m,b_1,b_2,\dots,b_m$ be nonnegative integers for which $n=\prod_{i=1}^mp_i^{a_i}$ and $k=\prod_{i=1}^mp_i^{b_i}$, where the $p_i$ are distinct primes. Then the existence of $s$ is equivalent to finding positive integers $c_1,c_2,\dots,c_m$ for which \[\prod_{i=1}^m\frac{a_i+c_i}{b_i+c_i}=1.\] Assume without loss of generality that $a_i<b_i$ if $1\le i\le u$, and $a_i>b_i$ otherwise. This is justified as the $\frac{a_i+c_i}{b_i+c_i}$ well always cancel if $a_i=b_i$ for some index $i$. Since $n\nmid k$ and $k\nmid n$, the value of $u$ is well-defined. Then for each $u\in\{2,\dots,m\}\setminus\{u+1\}$ set $c_i=c_{i-1}+b_{i-1}-a_i$. Set $X=b_1+\sum_{i=2}^u(b_i-a_i)$ and $Y=b_{u+1}+\sum_{i=2}^u(b_i-a_i)$, with $Y$ possibly negative. Then we have \[\prod_{i=1}^m\frac{a_i+c_i}{b_i+c_i}=\left(\frac{a_1+c_1}{X+c_1}\right)\left(\frac{a_{u+1}+c_{u+1}}{Y+c_{u+1}}\right).\]We have $a_1<X$ and $Y<a_{u+1}$, and setting this to $1$ reduces to an equation that admits an infinite number of solutions across the positive integers, as desired. $\blacksquare$

didnt solve it