2025 International Mathematical Olympiad China National Team Selection Test

Proof: We first prove a lemma. Lemma: For integers $m\ge 2$, $s\ge 1$, and a sequence $1=x_1<\cdots <x_s$ where every sum ${\sum }_{1\le j\le m}{x}_{i_j}$ ($1\le i_1\le \cdots \le i_m\le s$) is square-free, there exists an integer $x>x_s$ such that: $x$ is coprime with each $x_i$ ($1\le i\le s$) Every sum $(m-t)x+{\sum }_{1\le j\le t}{x}_{i_j}$ ($0\le t\le m-1$, $1\le i_1\le \cdots \le i_t\le s$) is square-free Proof of Lemma: Take $B>1$ such that: $$c_B:=\prod _{p<B}{\left(1-\frac{1}{p}\right)}^{-1}>m\cdot 2^{m+s}.$$ Let $$d_B=\prod _{p\le B}p,M_0=\prod _{1\le k\le 2^{m+s}{x}_s}k,M=d_B{M}_0.$$ Consider $x=y{M}^2+x_1$ ($1\le y\le M^2$). Clearly, $x$ is coprime with each $x_i$ ($1\le i\le s$). Note that $$(m-t)x+\sum _{1\le j\le t}{x}_{i_j}\le mx\le m(M^4+1)<(m{M}^2{)}^2.$$ If $q^2\mid (m-t)x+{\sum }_{1\le j\le t}{x}_{i_j}$, then $q<m{M}^2$. If $q\nmid M$, clearly $(m-t)x+{\sum }_{1\le j\le t}{x}_{i_j}$ is not divisible by $q^2$. For primes $q<m{M}^2$ with $q\nmid M$ (so $q>2^{m+s}{x}_s$), let $X_q$ be the set of integers $1\le y\le M^2$ where some sum $(m-t)x+{\sum }_{1\le j\le t}{x}_{i_j}$ is divisible by $q^2$. The number of possible index sets is at most $2^{s+m-2}$, so $$|X_q|\le 2^{m+s-2}\left(\frac{M^2}{q^2}+1\right).$$ For $2^{m+s}{x}_s<q\le M$: $$|X_q|\le 2^{m+s-1}\frac{M^2}{q^2};$$ ---

For $M<q\le m{M}^2$: $$|X_q|\le 2^{m+s-1}.$$ Thus, $$\begin{array}{rl}|\bigcup X_q|& \le 2^{m+s-1}{M}^2\sum _{q>2^{m+s}{x}_s}\frac{1}{q^2}+2^{m+s-1}(\pi (m{M}^2)-\pi (M))\\ & \le \frac{2^{m+s-1}}{2^{m+s}{x}_s}{M}^2+\frac{m\cdot 2^{m+s-1}}{c_B}{M}^2<M^2.\end{array}$$ Since we have used $d_B\mid M$, it follows that $$\pi (m{M}^2)-\pi (M)\le (m{M}^2-M)\prod _{p\le B}\left(1-\frac{1}{p}\right)<c_B^{-1}m{M}^2.$$ Now choose an integer $y$ such that $1\le y\le M$ and $$y\notin \bigcup _{\begin{array}{c}q<m{M}^2\\ q\nmid M\end{array}}{X}_q,$$ then $x=y{M}^2+x_1$ will satisfy the requirements. Now the main proof. If $m$ has a square factor $p^2$, by pigeonhole principle there exist $m$ elements $a_{i_1},\dots ,a_{i_m}$ with $a_{i_1}\equiv \cdots \equiv a_{i_m}(\mathrm{mod}{p}^2)$, making their sum divisible by $p^2$. If $m$ is square-free, construct $A=\{a_i\}$ inductively starting with $a_1=1$. By the lemma, we can find pairwise coprime integers $1=a_1<a_2<\dots$ where all $m$-term sums are square-free. Then all $m$-term products are also square-free since the $a_i$ are pairwise coprime. $\square$