Ehh INMO was trash. I think I will get 17/0/0/0-1/3-5/10-14, which is def not good enough for qualifying from 12th grade. Well, I really feel sad but let's not talk about it and focus on EGMO rather. INMO 2023 P1 Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square. I will use Atul's sol, cause it's the exact same as mine. Proof: Consider the graph $G$ induced by the elements of $S$ and edges being if the products are perfect squares. Note that if $xy = a^2$ and $xz = b^2$, then $yz = \left( \frac{ab}{x} \right)^2$, since its an integer and square of a rational number its a perfect square and so $yz$ is an edge too. So the graph is a bunch of disjoint cliques, say with sizes $c_1, c_2, \cdots, c_k$. Then $\sum_{i=1}^k c_i^2 = 2023$, which ...
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