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勾股方程与二次剩余

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孙智伟教授猜测:对于每个奇素数p>100,可要求勾股方程x2+y2=z2的解x,y,z ∈[1,p],且分别为模p的二次剩余或者二次非剩余(共八种情形).对于所有充分大的素数p,本文证明了这一猜测,其方法主要涉及Lillian B.Pierce和Junyan Xu所证明的关于多元高次型的特征和的Burgess界.
Pythagoras Equations and Quadratic Residues
It is conjectured by Professor Zhi-Wei Sun that for each given odd prime p>100,there always exists an solution(x,y,z)∈[1,p]3 to the Pythagoras equation x2+y2=z2 such that x,y,z are quadratic residues or non-residues modulo p respec-tively(eight cases in total).In this paper,we are able to prove the above assertion for all sufficiently large primes p,and the method is based on the recent Burgess bound for character sums of forms in many variables due to Lillian B.Pierce and Junyan Xu.

Pythagoras equationquadratic residuecharacter sumBurgess bound

郗平、郑钧仁

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西安交通大学数学与统计学院 西安 710049

勾股方程 二次剩余 特征和 Burgess界

国家自然科学基金资助项目国家自然科学基金资助项目

1197137012025106

2024

数学学报
中国科学院数学与系统科学研究院数学研究所

数学学报

CSTPCD北大核心
影响因子:0.261
ISSN:0583-1431
年,卷(期):2024.67(2)
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