论文标题
使用线性复发和数值分析，在不公平的0-1多项式猜想上的进展
Progress on the unfair 0-1-polynomials conjecture using linear recurrences and numerical analysis
论文作者
论文摘要
如果两个具有实际非负系数的一元多项式的乘积具有等于0或1的所有系数，那么这两个因素的所有系数也等于0或1？这是这个有趣的问题的等效表述：是否可以不公平地称重一对骰子，以便每个可能的结果的概率（滚动并拿出总和）是相同的吗？如果两个骰子的六个面孔编号为1到6，则很容易证明答案是否定的。但是对于有限多面孔的一般骰子，这是一个空旷的问题，自1937年以来就没有明显的进步。在本文中，我们研究了第一个无法用经典方法处理的第一个无限案件家族：第一个模具的面孔为0,2和5，而第二个则是第二个模具，而第二个则是任意的。换句话说，我们检查了某些非负$ a $的0-1多项式的因素，其因子等于$ x^5 +a x^2 +1 $。我们发现，可以使用线性复发序列，计算结果，大量的分析和数值近似值以及一点点运气来解决此情况（即必定是$ a = 0 $或$ a = 1 $）。
If the product of two monic polynomials with real nonnegative coefficients has all coefficients equal to 0 or 1, does it follow that all the coefficients of the two factors are also equal to 0 or 1? Here is an equivalent formulation of this intriguing problem: is it possible to weigh unfairly a pair of dice so that the probabilities of every possible outcome (roll them and take the sum) were the same? If the two dice have six faces numbered 1 to 6, it is easy to show that the answer is no. But for general dice with finitely many faces, this is an open problem with no significant advancement since 1937. In this paper we examine, in some sense, the first infinite family of cases that cannot be treated with classical methods: the first die has three faces numbered with 0,2 and 5, while the second die is arbitrary. In other words, we examine factorizations of 0-1-polynomials with one factor equal to $x^5+a x^2 +1$ for some nonnegative $a$. We discover that this case may be solved (that is, necessarily $a=0$ or $a=1$) using the theory of linear recurrence sequences, computation of resultants, a fair amount of analytic and numerical approximations... and a little bit of luck.
