China National Team Selection Test

We first prove the following lemma:

Lemma. Let $u$ be an integer with $a\le u<u+1<b$. Then there are two distinct integers $x$, $y$ in the interval $[ab,(a+1)(b+1))$, such that $\frac{xy}{u(u+1)}$ is a square of an integer.

Proof of lemma. Let $v$ be the smallest integer not less than $\frac{ab}{u}$, i.e. $v$ satisfies $$\frac{ab}{u}\le v<\frac{ab}{u}+1;$$ hence $$ab\le uv<ab+u(ab+a+b+1),\stackrel{◯}{1}$$ and thus $$ab<(u+1)v=uv+v<ab+u+\frac{ab}{u}+1<ab+a+b+1\text{ (since }a\le u<b\text{).}\stackrel{◯}{2}$$ (Here we have used a well-known result: the function $f(t)=t+\frac{ab}{t}$ ($a\le t\le b$) attains its maximum at $t=a$ or $b$.)

By ① and ②, we see that $uv$ and $(u+1)v$ are two distinct integers in the interval $I=[ab,(a+1)(b+1))$. Let $x=uv$ and $y=(u+1)v$. Then $\frac{xy}{u(u+1)}=v^2$ is a square of an integer number. We have verified the lemma.

Going back to the original problem, suppose that $m<n$. Then $a\le m\le n-1<b$. It follows from the lemma that for every $k=m,m+1,\dots ,n-1$, there exist $x_k$, $y_k$, two distinct integers in the interval $[ab,(a+1)(b+1))$, and integer $A_k$, such that $$\frac{x_k{y}_k}{k(k+1)}=A_k^2.$$ Multiplying all together, we find that $$\frac{{\prod }_{k=m}^{n-1}{x}_k{y}_k}{mn(m+1{)}^2\cdots (n-1{)}^2}=\prod _{k=m}^{n-1}{A}_k^2$$ is a square of an integer.

Let $S$ be the set of numbers that appear in $x_i$, $y_i$ ($m\le i\le n-1$) odd times. If $S$ is nonempty, then it follows from the above equality that $\frac{\prod x_i}{mn}$ is a square of a rational number.

If $S$ is empty, then $mn$ is a square of an integer. Since $a+b>2\sqrt{ab}$, we have $ab+a+b+1>ab+2\sqrt{ab}+1$, i.e. $\sqrt{(a+1)(b+1)}>\sqrt{ab}+1$, which means that there is at least an integer in the interval $[\sqrt{ab},\sqrt{(a+1)(b+1)})$. As a result there is a perfect square in the interval $[ab,(a+1)(b+1))$. Suppose that $r^2\in [ab,(a+1)(b+1))$ ($r\in \mathbb{Z}$), and let $S^{\prime }=\{r^2\}$. Then $\frac{{\prod }_{x\in S^{\prime }}x}{mn}$ is a square of a rational number.