是否存在三个长度为偶数的区间，两两交集长度均为奇数？

We can prove this using a straightforward parity argument that exposes a contradiction in the requirements. Here's the breakdown:

Step 1: Model intervals with parity #

Each interval (I_i = [a_i, b_i]) has even length, so (b_i - a_i \equiv 0 \pmod{2}). This means the start and end of every interval share the same parity: (a_i \equiv b_i \pmod{2}). Let’s define (p_i = a_i \mod 2) (0 for even, 1 for odd) — this parity applies to both endpoints of (I_i).

Step 2: Parity of intersection length #

For two overlapping intervals (I_i) and (I_j), their intersection length is (\min(b_i, b_j) - \max(a_i, a_j)). Calculating this modulo 2:

• (\min(b_i, b_j) \equiv \min(p_i, p_j) \pmod{2}) (since (b_i) matches (p_i)'s parity)
• (\max(a_i, a_j) \equiv \max(p_i, p_j) \pmod{2}) (since (a_i) matches (p_i)'s parity)

The difference modulo 2 simplifies to:
[
\min(p_i, p_j) - \max(p_i, p_j) \pmod{2}
]
Testing all parity pairs:

• If (p_i = p_j): The difference is (0-0=0) or (1-1=0), so even length.
• If (p_i \neq p_j): The difference is (0-1=-1 \equiv 1 \pmod{2}), so odd length.

Step 3: The contradiction #

To satisfy all three odd-length intersections, we’d need:

• (p_1 \neq p_2) (for (I_1 \cap I_2) to be odd)
• (p_2 \neq p_3) (for (I_2 \cap I_3) to be odd)
• (p_1 \neq p_3) (for (I_3 \cap I_1) to be odd)

But parities are binary (only 0 or 1). If (p_1 \neq p_2) and (p_2 \neq p_3), then (p_1) must equal (p_3) (there’s no third parity value). This means (I_1 \cap I_3) would have even length, violating the requirement.

Final Conclusion #

It’s mathematically impossible to have three even-length intervals where all pairwise intersections have odd length. The parity constraints create an unavoidable contradiction.

内容的提问来源于stack exchange，提问作者Savannah