Does anyone have a solution? Also what is the difficulty of IRAN 3rd round is assumed to be?(Is it IRAN TST level ,IMO level or something else?) Thanks!!!

In my opinion, Iran 3rd round is much easier than IMO (maybe a bit harder than USAJMO).

Similar to this https://artofproblemsolving.com/community/c6h1512905_set_on_naturals

Are $x,y$ given integers? Or are they arbitrarily chosen?

lminsl wrote: Are $x,y$ given integers? Or are they arbitrarily chosen? The final answer depends on the values of $x,y$.

tenplusten wrote: Does anyone have a solution? Also what is the difficulty of IRAN 3rd round is assumed to be?(Is it IRAN TST level ,IMO level or something else?) Thanks!!! that is so easy

Thanks below,the answer should be: if $v_2(x)=v_2(y)$,then the minimalist is 2, otherwise the answer is 3.

liekkas wrote: WLOG $x < y$, then the answer is $3$ when $y=(2k+1)x < n^2$, where $k$ is an integer, and $2$ otherwise. Take n=3, and x=2, y=3, you'll find a contradiction.

omarius wrote: Take n=3, and x=2, y=2 x,y must be distinct.

Clearly we require at least $2$ colours. If $v_2(x)=v_2(y)=k$ then we write $x=2^k \cdot x',y=2^k \cdot y'$. We colour the tape alternating $\text{red, blue}$ in blocks of $2^k$. Assume for contradiction two $\text{red}$ squares are separated by distance $x$ as they occur in blocks of $2^k$ considering $\pmod{2^k}$ the squares must have the same residue but then the blocks are separated by even multiples of $2^k$ meaning $2 \cdot 2^k=2^{k+1} \vert x$ which is a contradiction. We can do the same for $y$ so this works. If $k=v_2(x) <v_2(y)$ then if we only have $2$ colours consider label $1+\frac{xy}{2^k}$ this differs from label $1$ by an even multiple of $x$ but an odd multiple of $y$ and as the colouring must alternate this yields a contradiction. Now we piecewise construct the colouring going from labels $1,2,\cdots$. At each point we cannot choose at most $2$ colours so we can always choose an allowed colour. This ensures the colouring is valid.

Case $1 : x,y$ are odd. For every cell $t$ if $t \equiv 0$ $(mod 2)$ color it $A$ and else color it $B$ and it clearly works. Case $2 : v_2(x) = v_2(y) = k$. Let $S_t$ be set of numbers which for every $a \in S_t$ we have $a \equiv t$ $(mod 2^k)$. Note that $\frac{x}{2^k}$ and $\frac{y}{2^k}$ are odd so in each set we color them same as last case. Case $3 : v_2(x) \neq v_2(y)$. Let $M = lcd(x,y)$ Note that since $M \le xy \le n(n-1)$, number $M +1$ exists and Note that $v_2(\frac{M}{x}) \neq v_2(\frac{M}{y})$ so by parity we can't use only $2$ colors but $3$ is ok since for every cell $t$ in tape, we have $3$ color options which at most $2$ of them are not available.