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Arithmetic properties of polynomial solutions of the Diophantine equation $$P(x)x^{n+1}+Q(x)(x+1)^{n+1}=1$$ P ( x ) x n + 1 + Q ( x ) ( x + 1 ) n + 1 = 1
Periodica Mathematica Hungarica ( IF 0.6 ) Pub Date : 2021-04-29 , DOI: 10.1007/s10998-020-00376-5
Karl Dilcher , Maciej Ulas

For each integer \(n\ge 1\) we consider the unique polynomials \(P, Q\in {\mathbb {Q}}[x]\) of smallest degree n that are solutions of the equation \(P(x)x^{n+1}+Q(x)(x+1)^{n+1}=1\). We derive numerous properties of these polynomials and their derivatives, including explicit expansions, differential equations, recurrence relations, generating functions, resultants, discriminants, and irreducibility results. We also consider some related polynomials and their properties.



中文翻译:

Diophantine方程$$ P(x)x ^ {n + 1} + Q(x)(x + 1)^ {n + 1} = 1 $$ P(x)xn + 1 +的多项式解的算术性质Q(x)(x + 1)n + 1 = 1

对于每个整数\(n \ ge 1 \),我们考虑最小阶数n的唯一多项式\(P,Q \ in {\ mathbb {Q}} [x] \)是方程\(P(x )x ^ {n + 1} + Q(x)(x + 1)^ {n + 1} = 1 \)。我们推导了这些多项式及其导数的许多属性,包括显式展开,微分方程,递归关系,生成函数,结果,判别和不可约结果。我们还考虑了一些相关的多项式及其性质。

更新日期:2021-04-30
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