超椭圆曲线yk = x(x + 2m) 上的有理点

设k ≥ 2; m ≥ 0 是整数, 给出了超椭圆曲线y^k = x(x + 2^m) 上所有的有理点(x, y).     

椭圆曲线y~2=(x+6)(x~2-6x+23)的正整数点

利用初等方法证明了椭圆曲线y~2=(x+6)(x~2-6x+23)无正整数点,研究结果对于a,p∈Z时,椭圆曲线y~2=(x+a)(x~2-ax+p)的求解有一定的借鉴作用,同时此结果推进了该类椭圆曲线的研究.     

椭圆曲线y~2=(x+2)(x~2-2x+43)的整数点

利用初等方法证明了椭圆曲线y~2=(x+2)(x~2-2x+43)仅有整数点(x,y)=(-2,0).     

椭圆曲线y~2=(x+2)(x~2-2x+15)的正整数点

利用初等方法证明了椭圆曲线y~2=(x+2)(x~2-2x+15)无正整数点.     

关于不定方程y(y+1)(y+2)(y+3)=4n~2x(x+1)(x+2)(x+3)

设1n∈N*,运用Pell方程的一些结果以及代数数论和p-adic分析方法证明了不定方程y(y+1)(y+2)(y+3)=4n~2x(x+1)(+2)(x+3)(x,y∈N*)除开n=1189时仅有一组解(x,y)=(33,1680)外,无其他解.     

椭圆曲线y2=(x+p)(x2+p2)的本原整数点

设p是奇素数.本文给出了椭圆曲线y2=(x+p)(x2+p2)存在可使y为偶数的本原整数点(x,y)的充要条件.     

关于不定方程3x(x+1)(x+2)(x+3)=10y(y+1)(y+2)(y+3)

通过利用pell方程、递归序列、平方剩余、Legendre符号、同余关系等初等证明方法,并利用Mathematica软件对Legendre符号等进行计算,证明了方程3x(x+1)(x+2)(x+3)=10y(y+1)(y+2)(y+3)共有16组整数解,并且无正整数解.     

从(x±1)(x~2±x+1)=x~3±1谈起

乘法公式中有 (x+1)(x~2-x+1)=x~3+1,(x-1)(x~2+x+1)=x~3-1。等式两边互换,就得到因式分解 x~3+1=(x+1)(x~2-x+1),x~3-1=(x-1)(x~2+x+1)。进而有 x~4+1=(x+1)(x~3-x~2+x-1),x~4-1=(x-1)(x~3+x~2+x+1)。推广这些公式,可以得到定理1 (1)对任意正整数n,有 x~n-1=(x-1)(x~(n-1)+x~(n-2)+…+x+1)     

椭圆曲线y~2=x(x-p)(x-q)的整数点(Ⅰ)

设p,q为奇素数,m为正奇数,且p+2~m=q,p≡3(mod4).证明了:当m=1或3时,椭圆曲线y~2=x(x-p)(x-q)(xq)至多有1对整数点(x,y);当m≥5时,该椭圆曲线至多有2对整数点(x,y).同时具体给出了(p,q)=(71,103)时椭圆曲线的全部整数点.     

关于丟番图方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)

先后运用了pell方程、勒让德符号,同余关系,递归序列、二次平方剩余,分类讨论的有关方法,并通过使用数学软件Mathematica进行计算,证明了以下结论:不定方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)没有正整数解,并找出了该方程的全部16组整数解.     

Über die Gleichung (x + y)(x 2 +y 2)

Ohne Zusammenfassung     

On x 6 + x + a in Characteristic Three

Let F be a field of characteristic 3 and 0 a F. We show that the 10 ways to factor x ⁶ + x + a into two cubics over the algebraic closure F are in natural Galois bijection with the 10 roots of x ¹⁰ + ^ax + 1. We use this to (1) prove the two polynomials have the same splitting field; (2) prove that a difference set constructed by Arasu and Player using the polynomial x ⁶ + x + a is isomorphic to a difference set constructed by Dillon using the polynomial x ¹⁰ + x + a; (3) obtain a natural realization for the accidental isomorphism between the alternating group A ₆ and the special linear group PSL₂(9); and (4) characterize how x ⁶ + x + a factors when F = GF (3^m) with m odd. For example, x ⁶ + x + a is irreducible if and only if a can be written as ^{– 36} + ⁴ with F ^× and Tr(⁵) 0.     

方程a2(x)(t-τ)+a1(x)(t-τ)+a0x(t-τ)+b2(x)(t)+b1(x)(t)+b0x(t)=δ的部分解讨论

讨论了方程a2(x)(t-τ)+a1(x)(t-τ)+a0x(t-τ)+b2(x)(t)+b1(x)(t)+b0x(t)=δ的部分解.     

Calculation of coupling factors for the equation x2ϕ″−(x3+a2x2+a1x+a0)ϕ

A numerical realization of the method of finding coupling factors, described in the author's previous papers, is given in the present paper and its efficiency is studied. It is shown that this method is a generalization of the traditional method of finding coupling factors for Bessel functions. Several relations containing coupling factors are also derived. In a number of cases it is possible to judge the calculation error from the accuracy with which they are satisfied.Translated from Zapiski Nauchnykh Seminarov Leningradskogo Otdeleniya Matematicheskogo Instituta im. V. A. Steklova AN SSSR, Vol. 156, pp. 109–124, 1986.     

On the functional equation x + f(y + f(x))=y + f(x + f(y))

For an abelian group (G, + ,0) we consider the functional equation $$f : G \to G, x + f(y + f(x)) = y + f(x + f(y)) \quad (\forall x, y \in G), \quad\quad\qquad (1)$$ most times together with the condition $$f(0) = 0.\qquad\qquad\qquad\qquad\qquad (0)$$ Our main question is whether a solution of ${(1) \wedge (0)}$ must be additive, i.e., an endomorphism of G. We shall answer this question in the negative (Example 3.14) Rätz (Aequationes Math 81:300, 2011).     

Weitere Untersuchungen über die ternäre biquadratische Formf =x 1 3 x 2+x 2 3 x 3+x 3 3 x 1

Ohne Zusammenfassung