Problem 1605

1. Define the relation and intervals:
Let $m \sim n$ be the relation such that $\frac{\prod_{x \in S} x}{mn}$ is a rational square. We denote $[ab, (a+1)(b+1))$ as the "big interval" and $[a, b)$ as the "little interval".

2. Prove reflexivity:
Reflexivity requires showing that $m \sim m$ for any $m$ in the little interval. This means there exists a perfect square in the big interval. Consider $\sqrt{ab} + r$ where $0 \le r < 1$ such that $\sqrt{ab} + r$ is an integer. We need to show:
$$(\sqrt{ab} + r)^2 < (a+1)(b+1)$$
Expanding and simplifying:
$$(\sqrt{ab} + r)^2 = ab + r^2 + 2r\sqrt{ab}$$
Since $0 \le r < 1$:
$$ab + r^2 + 2r\sqrt{ab} < ab + 1 + 2\sqrt{ab}$$
Given $b - a > 2$, we have:
$$ab + 1 + 2\sqrt{ab} < ab + a + b + 1 = (a+1)(b+1)$$
Thus, there exists a perfect square in the big interval, proving reflexivity.

3. Prove symmetry:
Symmetry is immediate because if $m \sim n$, then $n \sim m$ by the definition of the relation.

4. Prove transitivity:
Assume $m \sim n$ and $n \sim p$. Let $S_1$ be the set associated with $m \sim n$ and $S_2$ be the set associated with $n \sim p$. We need to show $m \sim p$. Consider the symmetric difference $S_1 \oplus S_2$:
$$\frac{\prod_{x \in S_1 \oplus S_2} x}{mp}$$
If $S_1 \oplus S_2$ is empty, then $mp$ is a perfect square. Otherwise, $S_1 \oplus S_2$ provides the required set, proving transitivity.

5. Show all elements are equivalent:
We need to show $x \sim (x+1)$ for all integers $x$ in the little interval. This means finding $y$ such that:
$$ab \le xy \le (x+1)y < (a+1)(b+1)$$
We need to ensure:
$$xy \ge ab \quad \text{and} \quad (x+1)y < (a+1)(b+1)$$
If $xy < ab$ and $(x+1)(y+1) \ge (a+1)(b+1)$, subtracting these gives:
$$x + y > a + b$$
Since $a \le x \le y < b$, this is impossible. Therefore, such $y$ exists, proving $x \sim (x+1)$.

6. Conclusion:
Since $x \sim (x+1)$ for all $x$ in the little interval, all elements in the little interval are equivalent to each other.

$\blacksquare$