I will replace $2020$ by $N$ and prove it for arbitrary $N$. More concretely, I will show there exists a $\mathcal{C} = \{a_i:1\le i\le N\}$ enjoying the property. To that end, consider two lists of $\binom{N}{2}+N$ pairwise distinct primes, where the lists are $\{p_{ij}:1\le i<j\le N\}$ and $\{q_i:1\le i\le N\}$. Make the following assignment: for each $1\le i\le N$, $q_i$ divides only $a_i$. For each $1\le i<j\le N$, $p_{ij}$ divides $a_i$ and $a_j$, but no other numbers from the list. We now show this set satisfies the conditions. Clearly all numbers are distinct, as they have a prime factor ($q_i$) not contained in any other number. Now, for the gcd, fix a pair $1\le i<j\le N$, and note that $d_{ij}\triangleq {\rm gcd}(d_i,d_j)$ is divisible by $p_{ij}$. Now, fix another pair $1\le i'<j'\le N$ where $\{i',j'\} \ne \{i,j\}$. Note that if $p_{ij}\mid d_{i'j'}$, then $p_{ij}$ divides at least three distinct numbers from the list, not possible. Thus $d_{ij}$ are all distinct. Likewise, let $M_{ij}={\rm lcm}[a_i,a_j]$. For any distinct pair $(i,j)$ and $(i',j')$ if they have zero intersection, then $q_i$ divides $M_{ij}$ but not $M_{i'j'}$, and if they have an intersection of one, say at $i$, then $j\ne i'$ and therefore $q_j \mid M_{ij}$ but $q_j\nmid M_{i'j'}$. This concludes the construction.

$\prod_{i=1}^{2020}p_i,\prod_{j=2}^{2021}p_j,\prod_{k=3}^{2022}p_i\ldots \prod_{n=2020}^{4039}p_n$ should work, where $p_x$ is the xth smallest prime number

Let $T_i(a_{i,1},\cdots,a_{i,2020})$ be the representation vectorial of the power of the first $2020$ primes. So $$T_1=(0,1,\cdots,2019)$$$$T_2=(2019,0,1,\cdots, 2018)$$$$T_3=(2018,2019,0,\cdots, 2017)$$$$\vdots$$Satisfy the condition

Let $\{a\}_{n=0}^{2019}$ be the sequence of such numbers that they satisfy the condition of the problem. We set $a_i=t_i.p_1^{2^{2019-i}}$,where $t_i$ is the number such that it has $2^{i}-1$ prime divisors, so we can set $t_i=\prod_{k=1}^{2^{i}-1}p_{1+k}$ (here $t_0=1$). By checking we see that this satisfies the conditions of the problem.

mathleticguyyy wrote: $\prod_{i=1}^{2020}p_i,\prod_{j=2}^{2021}p_j,\prod_{k=3}^{2022}p_i\ldots \prod_{n=2020}^{4039}p_n$ should work, where $p_x$ is the xth smallest prime number I am fairly sure that this set works, but is there any way to rigorously prove this? I feel like this a little handwavy. I've always had some trouble trying to prove these types of constructions "work". Could someone explain?

\bump could anyone explain?

JustKeepRunning wrote: I am fairly sure that this set works, but is there any way to rigorously prove this? I feel like this a little handwavy. I've always had some trouble trying to prove these types of constructions "work". Could someone explain? For $1 \leq k < \ell \leq 2020$, $\text{lcm} \left( \prod_{i=k}^{k+2019}p_i, \prod_{j=\ell}^{\ell+2019}p_j \right) = \prod_{i=k}^{\ell +2019}p_i$ (this is since $k+2019$ and $\ell$ "overlap", as in $\ell \leq k + 2019$). For the same reason $\text{gcd} \left( \prod_{i=k}^{k+2019}p_i, \prod_{j=\ell}^{\ell+2019}p_j \right) = \prod_{i=\ell}^{k +2019}p_i$. Clearly both of these are unique for each pair $(k, \ell)$. An alternate construction would be the following: Let $P$ be the product of the first $2020$ primes ($P = p_1p_2, \cdots, p_{2020}$), and let $q_1, q_2, \cdots, q_{2020}$ be $2020$ primes different from those that divide $P$. Then, $\mathcal{C} = \{\frac{q_1P}{p_1}, \frac{q_2P}{p_2}, \cdots , \frac{q_{2020}P}{p_{2020}} \}$ works.

Al3jandro0000 wrote: Let $T_i(a_{i,1},\cdots,a_{i,2020})$ be the representation vectorial of the power of the first $2020$ primes. So $$T_1=(0,1,\cdots,2019)$$$$T_2=(2019,0,1,\cdots, 2018)$$$$T_3=(2018,2019,0,\cdots, 2017)$$$$\vdots$$Satisfy the condition Someone asked for proof. A set of primes $P=\{p_1,p_2,\cdots,p_{2020}\}$ so the vector $(a_1,a_2,\cdots,a_{2020})=\prod_{i=1}^{2020} p_i^{a_i}$. Following the given numbers $T_i$ above we will show this indeed works. Notice clearly $p_i,p_j\nmid \gcd(T_i,T_j)$. Now suppose we have $lcm(T_m,T_n)=lcm(T_x,T_y)$. So we have $$\frac{T_mT_n}{\gcd(T_m,T_n)}=\frac{T_xT_y}{\gcd(T_x,T_y)}$$So $$T_mT_n\gcd(T_x,T_y)=T_xT_y\gcd(T_m,T_n)$$Notice $$\nu_{p_m}(T_n)+\nu_{p_m}(\gcd(T_x,T_y))=\nu_{p_m}(T_x)+\nu_{p_k}(T_y)$$so since $\nu_{p}(T_i)$ is exclusive by the cyclic form of $T_i$’s we have either $T_n=T_x$ or $T_n=T_y$, WLOG say $T_n=T_x$. So we have $$T_m\gcd(T_n,T_y)=T_y\gcd(T_n,T_m)$$Finally notice $$\nu_{p_n}(T_m)=\nu_{p_n}(T_y)$$so we have $T_m=T_n$ thus $lcm(T_m,T_n)=lcm(T_x,T_y)$ iff $(m,n)=(x,y)$.

Construction basing on induction: We'll prove a stronger claim through induction: "For each positive integer $n$, there exists $n$ distinct, positive integers $a_1,a_2, \ldots ,a_n$ that satisfies simultaneously the following properties: 1. When one computes the greatest common divisor of each pair of elements of $\mathcal{C}$, one gets a list of numbers that are all distinct. When one computes the least common multiple of each pair of elements of $\mathcal{C}$, one gets a list of numbers that are all distinct. 2. There exists $n$ distinct prime $p_1,p_2, \ldots ,p_n$ such that $p_i \mid a_i$ and $gcd(p_i,a_j)=1$ for all $1 \le 1 \le n, j \ne i$ The base case for $n=1,2,3,4$ can be checked easily. Now suppose we have the claim is done for all positive integer less than $n+1$, so our job now is to prove the case $n+1$: We're now choosing $N$ large enough such that $N > p_i$ for $ 1 \le i \le n$ and $gcd(N,p_i)=1$ Let $a_{n+1}=p_1p_2p_3 \ldots p_n$ Now easily compute that: \[\begin{array}{*{20}{l}} {\left\{ {\begin{array}{*{20}{l}} \begin{array}{l} \gcd ({a_{n + 1}},N{a_i}) = {p_i}\\ \\ \end{array}\\ {\gcd (N{a_i},N{a_j}) = N\gcd ({a_i},{a_j})} \end{array}} \right.\forall i = \overline {1,n} }\\ {}\\ {\left\{ {\begin{array}{*{20}{l}} \begin{array}{l} lcm({a_{n + 1}},N{a_i}) = \frac{{{a_{n + 1}}N{a_i}}}{{{p_i}}}\\ \\ \end{array}\\ {lcm(N{a_i},N{a_j}) = N.lcm({a_i},{a_j})} \end{array}} \right.\forall i = \overline {1,n} } \end{array}\] Which obviously satifies the condition 1, so now we have to convert the sequence to make it satify the condition 2 without losing the properties that we've just construct in condition 1. Considering $n+1$ distinct prime that haven't appeared before, let it be $q_1,q_2,q_3, \ldots ,q_{n+1}$. So now our final sequence of $n+1$ positive integers satifying both conditions are : \[{x_{n + 1}} = {q_{n + 1}}{a_{n + 1}};{x_i} = N{a_i}{q _i}\forall i = \overline {1,n} \] Hence we are done so far!!!

Start with $n$ number $(a_1,a_2,...,a_n)=(1,1,...,1)$. Then in each step choose $(i,j)$ and multiple $a_i,a_j$ by prime number $p_{i,j}$. Do these steps exactly once for each $(i,j)$ and choose new primes in each step, I mean $p_{i,j}\ne p_{s,t}$ if $\{i,j\}\ne \{s,t\}$. It's easy to see that the numbers we will get after $\binom {n}{2}$ steps satisfy both conditions.

Choose $a_i = 2^i*3^{2021-i}$