论文标题
蒙德里安难题：关于$ m（n）= 0 $ case的界限
The Mondrian Puzzle: A Bound Concerning the $M(n) = 0$ Case
论文作者
论文摘要
为了响应有关Mondrian难题的数字视频https://www.youtube.com/watch?v=49kvzriofB0，我们提供了一个小于给定阈值$ x $ x $ $ x $ $ m（n）\ neq 0 $ m（n）$ m（n）$ m（n）$ m（n $ m（n ney）的nimian nimian the mondrian nimian in Minim in Minim in Minim in Minim in Minim in Minim in Minim in Minim Inim nim nimian nim nim in Minim in Minim in Minim in Minim in Minim Inim nim nim in I. the mondry in I.在一套不一致的整数矩形中，最大，最小的矩形，$ n $ by $ n $ square。
In response to the Numberphile video regarding the Mondrian Puzzle https://www.youtube.com/watch?v=49KvZrioFB0, we provide a lower bound on how many integers less than a given threshold $x$ satisfy $M(n) \neq 0$ where $M(n)$ is the quantity in which the Mondrian Puzzle is interested, i.e. the minimal difference in area between the largest and smallest rectangle in a set of incongruent, integer-sided rectangles which tile an $n$ by $n$ square.
