Find the least natural number $n$ such that if the set $\{1,2,\dots,n\}$ is arbitrarily divided into two non intersecting subsets then one of the subsets contains three distinct numbers such that the product of two of them equals the third.
Lets say I have two sets:
$A$ and $B$
So $1,2,3$ have to be in one set, lets put them in $A$. This forces $6$ to be in $B$ If we put $4$ in A, then $8,12$ must be in $B$. If we put $5$ in $A$ then $10,15$ must be in $B$. If we put $7$ in $A$ then $14,21$ to be in $B$.
So right now I have:
$$ A=\{1,2,3,4,5,7\} $$ $$ B=\{6,8,12,10,14,15,21\} $$
I don't see a particular pattern so i am assuming there is a different approach to this problem because this could go on forever. Any ideas?
EDIT: I misunderstood the question. Here is my new sets $$ A=\{1,2,3,6,9,12,24,36,72,18\} $$ $$ B=\{4,5,7,8,32,10,40,56,28,35,20\} $$
Here is my set so far
The answer is $n=96$.
To prove this, take any $n\ge96$ and assume we partition the set $\{1,2,\cdots,n\}$ into the disjoint union of $A$ and $B$. Let $P(A)$ denote the set of products of distinct (and different from $1$) elements of $A$, similarly $P(B)$ denote the set of products of distinct (and different from $1$) elements of $B$.
Assume without loss of generality that $48\in B$ and consider cases as follows. Four cases are formed based on whether $2$ is in $A$ or in $B$, and whether $3$ is in $A$ or in $B$. In each case we assume that $A\cap P(A)=\emptyset$, $B\cap P(B)=\emptyset$, and derive a contradiction.
Case when $\{2,3\}\subseteq A$. Then $6\in B$ (since $6=2\cdot3$ and $\{2,3\}\subseteq A$). Then $8\in A$ (since $8=48/6$ and $\{6,48\}\subseteq B$), so $4=8/2\in B$. Then $24=3\cdot8=4\cdot6\in P(A)\cap P(B)$, which is enough to derive a contradiction. (Indeed, $24$ must be in either $A$ or $B$. In the former case the contradiction is that $24=3\cdot8\in A\cap P(A)$, in the latter case $24=4\cdot6\in B\cap P(B)$.)
It is perhaps easier to visualize the above argument as in the following table, where numbers further to the right are added to $A$ or $B$ as a consequence of numbers (at the left) that were added earlier.
$$ \begin{array}{l|c|c|c|c|r} \hline A & 2,\ {\color{red}3} & & {\color{red}8} & & {\color{red}{24}} &\\ \hline B & 48 & {\color{blue}6} & & {\color{blue}4} & {\color{blue}{24}} &\\ \hline \end{array} $$
Case when $2\in A$, $3\in B$. Then $16=48/3\in A$, $8=16/2\in B$, $\{6=48/8,\ 24=3\cdot8\}\subset A$, $\{4=24/6,\ 12=24/2,\ 48\}\subset B$, and $4\cdot12=48$. This case visualized as follows:
$$ \begin{array}{l|c|c|c|c|r} \hline A & 2 & 16 & & 6,\ 24 & &\\ \hline B & 3,\ {\color{blue}{48}} & & 8 & & {\color{blue}{4,\ 12}} &\\ \hline \end{array} $$
Case when $3\in A$, $2\in B$. Then $24=48/2\in A$, $8=24/3\in B$, $\{4=8/2,\ 6=48/8\}\subset A$, hence $4\cdot6=24$ with $\{4,6,24\}\subset A$. This case visualized as follows:
$$ \begin{array}{l|c|c|c|c|r} \hline A & 3 & {\color{red}{24}} & & {\color{red}{4,\ 6}} &\\ \hline B & 2,\ 48 & & 8 & & \\ \hline \end{array} $$
Finally, case $\{2,3\}\subseteq B$. Then $\{6=2\cdot3,\ 96=2\cdot48,\ 16=48/3\}\subset A$, a contradiction as $6\cdot16=96$. This last case visualized as follows:
$$ \begin{array}{l|c|c|c|c|r} \hline A & & {\color{red}{6,\ 96,\ 16}} & \\ \hline B & 2,\ 3,\ 48 & & \\ \hline \end{array} $$
It remains to show that we could partition the set $\{1,2,\cdots,95\}$ into the disjoint union $A\cup B$ such that no two distinct (and different from $1$) numbers in $A$ have a product in $A$, and no two distinct (and different from $1$) numbers in $B$ have a product in $B$.
That is, $A\cap P(A)=\emptyset$ and $B\cap P(B)=\emptyset$.
Using considerations as above, we start with $\{6,8,12,16,24,36,18\}\subset A$ and $\{2,3,4,48,72\}\subset B$, and add the remaining numbers up to $95$ one after the other in either $A$ or $B$ trying to avoid a conflict. This was done by hand (and after that checked with a computer).
The following partition works:
$A=\{6,8,10,12,14,15,16,18,20,21,22,24,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,50,51,52,55,57,58,62,63,65,68,69,74,75,76,77,78,82,85,86,87,91,92,93,94,95\}$
and
$B=\{1,2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,48,49,53,54,56,59,60,61,64,66,67,70,71,72,73,79,80,81,83,84,88,89,90\}.$
To make the verification easier we also list $P(A)\cap\{1,2,\cdots,96\}$ and $P(B)\cap\{1,2,\cdots,96\}$.
$P(A)\cap\{1,2,\cdots,96\}=\{48,60,72,80,84,90,96\}.$
$P(B)\cap\{1,2,\cdots,96\}=\{6,8,10,12,14,15,18,20,21,22,26,27,28,33,34,35,36,38,39,44,45,46,50,51,52,55,57,58,62,63,65,68,69,74,75,76,77,82,85,86,87,91,92,93,94,95,96\}.$
Somewhat arbitrary $1$ and all primes ended up in $B$. This partition is not unique as $1$, $11$, and all primes $p\ge17$ (or any subset of these) could be moved from $B$ to $A$ without harm. (But $13$ could not be moved from $B$ to $A$ as this would create a conflict $6\cdot13=78$.) Many other variations are likely possible too.
One last edit to put $P(A)$, $A$, $B$, $P(B)$ all together for easier visual inspecion.
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & & & & & & & & & & & & & & & \\ \hline A & & & & & &6 & & 8& &10 & &12 & &14 &15 & 16& \\ \hline B &1 &2 &3 &4 &5 & &7 & &9 & &11 & &13 & & & & \\ \hline P(B) & & & & & &6 & &8 & &10 & &12 & &14 &15 & & \\ \hline \end{array} $$
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & & & & & & & & & & & & & & & \\ \hline A & &18 & &20 &21 &22 & &24 & &26 &27 &28 & &30 & &32 & \\ \hline B &17 & &19 & & & &23 & &25 & & & &29 & &31 & & \\ \hline P(B) & &18 & &20 &21 &22 & & & &26 &27 &28 & & & & & \\ \hline \end{array} $$
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & & & & & & & & & & & & & &48 & \\ \hline A &33 &34 &35 &36 & &38 &39 &40 & &42 & &44 &45 &46 & & & \\ \hline B & & & & &37 & & & &41 & &43 & & & &47 &48 & \\ \hline P(B) &33 &34 &35 &36 & &38 &39 & & & & &44 &45 &46 & & & \\ \hline \end{array} $$
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & & & & & & & & & &60 & & & & & \\ \hline A & &50 &51 &52 & & &55 & &57 &58 & & & &62 &63 & & \\ \hline B &49 & & & &53 &54 & &56 & & &59 &60 &61 & & &64 & \\ \hline P(B) & &50 &51 &52 & & &55 & &57 &58 & & & &62 &63 & & \\ \hline \end{array} $$
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & & & & & &72 & & & & & & & &80 & \\ \hline A &65 & & &68 &69 & & & & &74 &75 &76 &77 &78 & & & \\ \hline B & &66 &67 & & &70 &71 &72 &73 & & & & & &79 &80 & \\ \hline P(B) &65 & & &68 &69 & & & & &74 &75 &76 &77 & & & & \\ \hline \end{array} $$
$$ \begin{array}{l|c|c|c|c|c|c|c|c|c|c|c|c|c|r} \hline P(A) & & & &84 & & & & & &90 & & & & & &96 & \\ \hline A & &82 & & &85 &86 &87 & & & &91 &92 &93 &94 &95 & & \\ \hline B &81 & &83 &84 & & & &88 &89 &90 & & & & & & & \\ \hline P(B) & &82 & & &85 &86 &87 & & & &91 &92 &93 &94 &95 &96 & \\ \hline \end{array} $$