关于丢番图方程X~3±1=1267y~2的整数解

关于丢番图方程x³±1=1267y3±1=1267y²的初等解法至今仍未解决.主要利用递归序列、同余式、平方剩余、Pell方程的解的性质、Maple小程序,证明了丢番图方程x2的初等解法至今仍未解决.主要利用递归序列、同余式、平方剩余、Pell方程的解的性质、Maple小程序,证明了丢番图方程x³-1=1267y3-1=1267y²有整数解(x,y)=(1,0),(60817,±421356),而丢番图方程x2有整数解(x,y)=(1,0),(60817,±421356),而丢番图方程x³+1=1267y3+1=1267y²仅有整数解(x,y)=(-1,0).     

非整边的直角三角形整距点问题

以直角顶点为原点 ,两直角边分别为 x轴和 y轴的正方向建立坐标系 .不妨设斜边所在直线方程为 ax +by=n,则方程 ax +by=n - kc(其中 a、b、c∈ N+,且 a2 +b2 =c2 ,k为整数 )的正整数解就是整距点的坐标 ,因此整距点问题与一类不定方程的正整数解联系起来 .设 a,b,n皆为正整数 ,有以下引理 .引理 1　方程 ax +by =n有整数解的充要条件是 (a,b) |n.引理 2　若 (a,b) =1,且 x0 ,y0 为方程 ax+by =n的一组解 ,则方程其它解可表示为 :x =x0 +bt,y =y0 - at(t为整数 ) .引理 3　设 (a,b) =1,则当 n>ab- a-b时 ,方程 ax +by =n必有非负整数解 .以…     

世界各地数学奥林匹克竞赛试题摘编

1　 (第 2 3届全俄中学奥林匹克竞赛试题 ,11年级 )求方程 (x2 - y2 ) 2 =1+ 16 y的整数解 .解　以下将证明方程(x2 - y2 ) 2 =1+ 16 y (1)的解是 (- 4,5 ) ,(4,5 ) ,(- 1,0 ) ,(1,0 ) ,(- 4,3) ,(4,3) .设x ,y是满足方程 (1)的两个整数 .注意到 ,若 y <0 ,则 1+ 16 y <0 ,则 1+ 16 y不是一个完全平方数 ;若 (x ,y)就是 (1)的解 .不失一般性 ,可设x≥ 0 .情形 1:若x≥y ,可令x =y +a且a∈N .方程 (1)可改写为 :4a2 y2 + 4 (a3- 4) y +a4 - 1=0 .故 y是二次方程 4a2 X2 + 4 (a3- 4)X +a4 - 1=0的一个解 .此时Δ =16 (- 8a3+a2 + 16 ) ,则一…     

关于阶乘的Erdös-Stewart猜想

设n是正整数;Po=1,p(i=1,2,…)是第i个素数.本文证明了方程n! +l=pαkP+1, pk-l＜n＜pk, a , b ∈Z, o＞0, b＞O,仅有解(n,pk,pk+1,a,6)=(1,1,2,1,0),(2,3,5,1,0),(3,5,7,O,1),(4,5,7,2,0),(5,7,11,O,2).上述结果证实了Erdos和Stewart提出的-个猜想.     

不定方程x4-y4=n整数解求法探讨

求方程 x4- y4=n　 ( n∈ N)的整数解 ,至今还没见到一般方法 ,本文将给出这类不定方程一种解法 .文中字母 P表示质数集 ,符号 ( a,b)( a、b∈ Z)表示不定方程　　 x4- y4=n　 ( n∈ N) ( 1 )的整数解 .定理 1　若 n∈ P,则方程 ( 1 )没有整数解 .证明　假定方程 ( 1 )有整数解 ( a,b) ,定有　 a2 b2 =n,　 a2 - b2 =1 ,∵　 a、b∈ Z,| a| >| b| ,只有　　　 (± 1 ) 2 - 0 2 =1 ,∴　 a =± 1 ,　 b =0 ,　 a2 b2 =1 ,与 a2 b2 =n是质数相矛盾 ,故方程 ( 1 )没有整数解 .由费马定理知 ,有定理 2　当 n =m4( n∈ N)时 ,则方程 ( 1…     

关于丢番图方程1+5~x+2~y5~z11~u=2~v·11~w,yvw>0,x+z>0的解

讨论了丢番图方程1+X+Y=Z的一个特殊情形.借助计算机,用初等方法给出了指数丢番图方程1+5~x+2~y5~z11~u=2~v·11~w,yvw>0,x+z>0的全部非负整数解.     

Diophantine方程x~2+2~m=y~n的全部解

运用高次Diophantine方程和指数Diophantine方程的己知结果证明了:方程x~2+2~m=y~n仅有正整数解(x,y,m,n)=(2~(3k)×5,2~(2k)×3,6k+1,3),(2~(2k)×7,2~k×3,4k+5,4),(2~(3k)×11,2~(2k)×5,6k+2,3),(2~(5k+2)×11,2~(2k+1)×3,10k+5,5),(2~(2kl+3k+l+1),2~(2k+1),4kl+6k+2l+2,2l+3),其中k和l是任意非负整数.     

关于不定方程ax+by=c的特解的算法

给定二元一次不定方程ａｘ+ｂｙ=ｃ ,( 1 )其中ａ ,ｂ ,ｃ为整数 ,且ａ ,ｂ≠ 0 .由文 [1 ]知 ,不定方程 ( 1 )有整数解的充分必要条件为ｇｃｄ(ａ ,ｂ) ｜ｃ,亦即ａ ,ｂ的最大公因数可整除ｃ.若不作特别说明 ,本文所说的解均指整数解 .设不定方程 ( 1 )有解 .不失一般性 ,我们总可假定ｇｃｄ(ａ ,ｂ) =1 .此时 ,不定方程 ( 1 )有解且它的全部解为ｘ=ｘ0 -ｂｔ,ｙ=ｙ0 +ａｔ,( 2 )其中 (ｘ0 ,ｙ0 )为不定方程 ( 1 )的特解 ,ｔ为任意整数 .这是因为若 ( 1 )的一般解为 (ｘ,ｙ) ,则 (ｘ -ｘ0 ,ｙ-ｙ0 )为 ( 1 )的齐次方程 (ｃ=0 )的一般解 .上…     

Mordell方程y~3=x~2+2p~4的正整数解

设p是奇素数.对于非负整数r,设U_(2r+1)=(α~(2r+1)+β~(2r+1))/2~(1/2),V_(2r+1)=(α~(2r+1)-β~(2r+1))/6~(1/2),其中α=(1+3~(1/2))/2~(1/2),β=(1-3~(1/2))/2~(1/2).运用初等数论方法证明了:方程y~3=x~2+2p~4有适合gcd(x,y)=1的正整数解(x,y)的充要条件是p=U_(2m+1),其中m是正整数.当上述条件成立时,方程仅有正整数解(x,y)=(V(2m+1)(V_(2m+1)~2-6),V_(2m+1)~2+2)适合gcd(x,y)=1.由此可知:当p10000时,方程仅有正整数解(p,x,y)=(5,9,11),(19,1265,123),(71,68675,1683)和(3691,9677201305,4541163)适合gcd(x,y)=1.     

关于阶乘的Erd(o)s-Stewart猜想

设n是正整数;Po=1,p(i=1,2,…)是第i个素数.本文证明了:方程n! +l=pαkP+1, pk-l＜n＜pk, a , b ∈Z, o＞0, b＞O,仅有解(n,pk,pk+1,a,6)=(1,1,2,1,0),(2,3,5,1,0),(3,5,7,O,1),(4,5,7,2,0),(5,7,11,O,2).上述结果证实了Erdos和Stewart提出的-个猜想.     

Quasi‐symmetric designs with fixed difference of block intersection numbers

The following results for proper quasi‐symmetric designs with non‐zero intersection numbers x,y and λ > 1 are proved.

• (1) Let D be a quasi‐symmetric design with z = y ? x and v ≥ 2k. If x ≥ 1 + z + z³ then λ < x + 1 + z + z³.
• (2) Let D be a quasi‐symmetric design with intersection numbers x, y and y ? x = 1. Then D is a design with parameters v = (1 + m) (2 + m)/2, b = (2 + m) (3 + m)/2, r = m + 3, k = m + 1, λ = 2, x = 1, y = 2 and m = 2,3,… or complement of one of these design or D is a design with parameters v = 5, b = 10, r = 6, k = 3, λ = 3, and x = 1, y = 2.
• (3) Let D be a triangle free quasi‐symmetric design with z = y ? x and v ≥ 2k, then x ≤ z + z².
• (4) For fixed z ≥ 1 there exist finitely many triangle free quasi‐symmetric designs non‐zero intersection numbers x, y = x + z.
• (5) There do not exist triangle free quasi‐symmetric designs with non‐zero intersection numbers x, y = x + 2.

© 2006 Wiley Periodicals, Inc. J Combin Designs 15: 49–60, 2007     

The generalised Fermat equation x 2 + y 3 = z 15

We determine the set of primitive integral solutions to the generalised Fermat equation x ² + y ³ = z ¹⁵. As expected, the only solutions are the trivial ones with xyz = 0 and the non-trivial one (x, y, z) = (± 3, ?2, 1).     

Oscillation of partial difference equations with continuous variables

In this paper, we consider the partial difference equation with continuous variables of the form P₁z(x + a, y + b) + p₂z (x + a, y) + p₃z (x, y + b) − p₄z (x, y) + P (x, y) z (x − τ, y − σ) = 0, where P ϵ C(R⁺ × R⁺, R⁺ − {0}), a, b, τ, σ are real numbers and p_i (i = 1, 2, 3, 4) are nonnegative constants. Some sufficient conditions for all solutions of this equation to be oscillatory are obtained.     

一类堆垒数论问题(Ⅰ)

§1 引言 堆垒数论中有一个著名的猜测: 设k_1,…,k_s是s个大于1的整数,且k_1≤k_2≤…≤k_s及sum from j=1 to s (k_j~(-1)>1)。又对每个素数p,对大的k,同余方程 X_1~(k_1)+X_2~(k_2)+…+X_s~(k_s)≡n(modp~k) 对某些j,P|x_j都有解,那末当n充分大时,方程     

An Excursion into Nonlinear Ramsey Theory

Formulas for two-color Rado numbers have been established for many families of linear equations. However, there are no explicit formulas for two-color Rado numbers for any nonlinear equations. In this paper, we will establish formulas for the two-color Rado numbers for three families of equations: x + y ⁿ = z, x + y ² + c = z, and x +  y ² = az , where c and a are positive integers.     

指数Diophantine方程4~x+b~y=(b+4)~2

本文研究了指数Diophantine方程4~x+b~y=(b+4)~2的解.设b1是给定的正奇数,运用有关指数Diophantine方程的已知结果以及有关Pell方程的Stormer定理的推广,证明了方程4~x+b~y=(b+4)~2仅有正整数解(x,y,z)=(1,1,1).     

On trigonometric functional equations of rectangular type

Summary Letf, G₁ × G₂ C, where G _i (i = 1, 2) denote arbitrary groups and C denotes the set of complex numbers. The general solutions of the following functional equationsf(x ₁ y ₁ ,x ₂ y ₂ ) +f(x ₁ y ₁ ,x ₂ y ₂ ^-1 ) +f(x ₁ y ₁ ^-1 ,x ₂ y ₂ ) +f(x ₁ y ₁ ^-1 ,x ₂ y ₂ ^-1 ) =f(x ₁ ,x ₂ )F(y ₁ ,y ₂ ) +F(x ₁ ,x ₂ )f(y ₁ ,y ₂ ) (1) andf(x ₁ y ₁ ,x ₂ y ₂ ) +f(x ₁ y ₁ ,x ₂ y ₂ ^-1 ) +f(x ₁ y ₁ ^-1 ,x ₂ y ₂ ) +f(x ₁ y ₁ ^-1 ,x ₂ y ₂ ^-1 ) =f(x ₁ ,x ₂ )f(y ₁ ,y ₂ ) +F(x ₁ ,x ₂ )F(y ₁ ,y ₂ ) (2) are determined assuming thatf satisfies the conditionf(x ₁y₁z₁, x₂) = f(x₁z₁y₁, x₂), f(x₁, x₂y₂z₂) = f(x₁, x₂z₂y₂) (C) for allx _i, y_i, x_i G_i (i = 1, 2). The functional equations (1) and (2) are generalizations of the well known rectangular type functional equationf(x ₁ + y₁, x₂ + y₂) + f(x₁ + y₁, x₂ – y₂) + f(x₁ – y₁, x₂ + y₂) + f(x₁ – y₁, x₂ – y₂) = 4f(x₁, x₂) studied by J. Aczel, H. Haruki, M. A. McKiernan and G. N. Sakovic in 1968.     

The exponential Diophantine equation x2 + (3n 2 + 1) y = (4n 2 + 1) z

Let n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x ² + (3n ² + 1) ^y = (4n ² + 1) ^z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n ³ + 3n, 1, 3).     

Three-variable Mahler measures and special values of modular and Dirichlet L-series

In this paper we prove that the Mahler measures of the Laurent polynomials (x+x ^?1)(y+y ^?1)(z+z ^?1)+k ^1/2, (x+x ^?1)²(y+y ^?1)²(1+z)³ z ^?2?k, and x ⁴+y ⁴+z ⁴+1+k ^1/4 xyz, for various values of k, are of the form r ₁ L′(f,0)+r ₂ L′(χ,?1), where $r_{1},r_{2}\in \mathbb{Q}$ , f is a CM newform of weight 3, and χ is a quadratic character. Since it has been proved that these Mahler measures can also be expressed in terms of logarithms and ₅ F ₄-hypergeometric series, we obtain several new hypergeometric evaluations and transformations from these results.