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
$$ab \le uv < ab + u(ab + a + b + 1), \quad \textcircled{1}$$
and thus
$$ab < (u+1)v = uv + v < ab + u + \frac{ab}{u} + 1 < ab + a + b + 1 \text{ (since } a \le u < b\text{).} \quad \textcircled{2}$$
(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 $S' = \{r^2\}$. Then $\frac{\prod_{x \in S'} x}{mn}$ is a square of a rational number.