IMO Shortlist 2021 C1

I am planning to do at least one ISL every day so that I do not lose my Olympiad touch (and also they are fun to think about!). Today, I tried the 2021 IMO shortlist C1.

(2021 ISL C1) Let

$S$ be an infinite set of positive integers, such that there exist four pairwise distinct

$a,b,c,d \in S$ with

$\gcd(a,b) \neq \gcd(c,d)$ . Prove that there exist three pairwise distinct

$x,y,z \in S$ such that

$\gcd(x,y)=\gcd(y,z) \neq \gcd(z,x)$ .

Suppose not. Then any

$3$ elements

$x,y,z\in S$ will be

$(x,y)=(y,z)=(x,z)$ or

$(x,y)\ne (y,z)\ne (x,z)$ .

There exists an infinite set

$T$ such that

$\forall x,y\in T,(x,y)=d,$ where

$d$ is constant.

Fix a random element

$a$ . Note that

$(x,a)|a$ . So

$(x,a)\le a$ .Since there are infinite elements and finite many possibilities for the gcd (atmost

$a$ ). So

$\exists$ set

$T$ which is infinite such that

$\forall b_1,b_2\in T$

$(a,b_1)=(a,b_2)=d.$ Note that if

$(b_1,b_2)\ne d$ then we get a contradiction as we get a set satisfying the problem statement. So in

$T,$

$\forall x,y\in T,(x,y)=d.$

Note that

$T\ne S$ . As, if

$T=S$ then we get all elements having same

$gcd$ which is a contradiction. So

$\exists b\in S$ such that

$(a,b)\ne d$ .Say

$d_2$ . But now consider

$T$ and

$b$ again. We get another infinite set

$R$ such that

$\forall r_1,r_2\in R,(r_1,b)=(r_2,b)=d_3.$ And

$R\subset T$ . Note that

$(r_1,r_2)=d$ Now, we give the following constructions: If

$d\ne d_3$ . Then note that

$(x,y,z)=(b,r_1,r_2)$ works. Contradiction. If

$d=d_3$ . Then note that

$(x,y,z)=(b,r_1,a)$ works. Contradiction. So we must have

$3$ elements

$x,y,z\in S$ such that

$(x,y)=(y,z)\ne(x,z)$ .

I think the main motivation was that infinite sets are too good to not use. The

$\gcd$ condition gives us the infinite sets. And then seeing how the infinite sets are actually an infinite clique. Really nice.

Will see you all tomorrow with a new combinatorics problem!

~Sunaina (Jan 12,2024)

Comments

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Sunaina thinks absurd

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