Non-Negative Integer Solutions of Two Diophantine Equations 2x + 9y = z2 and 5x + 9y = z2

In this paper, we study two Diophantine equations of the type px + 9y = z2 , where p is a prime number. We find that the equation 2x + 9y = z2 has exactly two solutions (x, y, z) in non-negative integer i.e., {(3, 0, 3),(4, 1, 5)} but 5x + 9y = z2 has no non-negative integer solution.

On Diophantine Equation X(X+1)(X+2)(X+3)=14Y(Y+1)(Y+2)(Y+3)

The Diophantine equation X（ X ＋ 1 ） （ X ＋ 2 ） （ X ＋ 3 ） = 14Y（ Y ＋ 1 ） （ Y ＋ 2 ） （ Y ＋ 3 ） still remains open. Using recurrence sequence, Maple software, Pell equation and quadraric residue, this paper proved it has only two positive integer solutions, i. e., （X,Y） = （5,2） ,（7,3）.

Heron Triangle and Diophantine Equation

In this paper, we study the quantic Diophantine equation （1） with elementary geometry method, therefore all positive integer solutions of the equation （1） are obtained, and existence of Heron triangle whose median lengths are all positive integer are discussed here.

A Variant of Fermat’s Diophantine Equation

A variant of Fermat’s last Diophantine equation is proposed by adjusting the number of terms in accord with the power of terms and a theorem describing the solubility conditions is stated. Numerically obtained primitive solutions are presented for several cases with number of terms equal to or greater than powers. Further, geometric representations of solutions for the second and third power equations are devised by recasting the general equation in a form with rational solutions less than unity. Finally, it is suggested to consider negative and complex integers in seeking solutions to Diophantine forms in general.

Diophantine equation x^2 + 2~m =y^n

LET Z,N,Q be the sets of integers, positive integers and rational numbers, respectively. The solutions (x, y, m, n) of the exponential Diophantine equation X^2+2~m=y^n,x,y,m,n∈N,2y,n【2 (1)are connected with many questions in number theory and combinatorial theory. In the recent fifty years, there were many papers concerned with the equation written by Ljunggren, Nagell, Brown, Toyoizumi and Cohn. In 1986, ref. [1] claimed that all solutions of (1) had been determined. However, we have not seen the proof so far. Therefore, the solution of (1) has not been found yet. In this note, using Baker’s method, we prove the following result.

三次Diophantine方程x^(3)+1=pQy^(2)的整数解

利用同余式、Legendre符号、递归数列、Pell方程解的性质以及一些初等数论方法得到以下结论:当p、Q分别为6k+1和6k-1型奇素数或p、Q为2个互不相同的6k+1型奇素数时,丢番图方程x^(3)+1=pQy^(2)仅有整数解(x,y)=(-1,0).

关于Diophantine方程(20n)~x+(21n)~y=(29n)~z(英文)

设a,b,c是满足条件a2+b2=c2的两两互素的正整数.Jesmanowicz于1956年猜想对于任意给定的正整数n,方程(an)x+(bn)y=(cn)z仅有解(x,y,z)=(2,2,2).本文证明了方程(20n)x+(21n)y=(29n)z有唯一解(x,y,z)=(2,2,2).

关于Diophantine方程x^3±1=pqy^2

关于Diophantine方程x3±1=Dy2至今仍未解决.论文利用同余式、平方剩余、Pell方程解的性质、递归序列证明:(1)p≡1(mod 12)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(1,0);(2)p≡1(mod 24)为素数,q=12s2+1(s是正奇数)为素数,(p q)=-1时,Diophantine方程x3±1=pqy2仅有整数解(x,y)=(-1,0).

关于Diophantine方程x^3±1=3Dy^2

设D是奇素数,运用同余式、平方剩余、递归序列、Maple程序等初等方法得出了当D=27t2+1(t∈Z+)时,Diophantine方程x3±1=3 Dy2无正整数解的一个充分条件.

关于Diophantine方程x^3-1=13qy^2的整数解

设D=multiply from i=1 to s p_i(s≥2),p_i=1(mod 6)(1≤i≤s)为不同的奇素数.关于Diophantine方程x^3-1=Dy^2的初等解法至今仍未解决.主要利用同余式、平方剩余、Pell方程的解的性质、递归序列,证明了q≡7(mod 24)为奇素数.(q/13)=-1时,Diophantine方程x^3-1=13qy^2仅有整数解(x,y)=(1,0).

关于三次Diophantine方程x^3+1=2p_1p_2Qy^2的可解性

设p_1,p_2是适合_p1≡p_2≡1(mod 6)以及(p_1/p_2)=-1的奇素数,其中(p_1/p_2)是Legendre符号。设Q是至少有两个不同素因数且每个素因数q都满足q≡5(mod 6)的无平方因子正整数。运用初等数论方法证明了:如果p_1≡1(mod 8),p_2≡5(mod 8),Q≡1(mod 4),那么方程x^3+1=2p_1p_2Qy^2无正整数解(x,y)。

关于指数Diophantine方程x^3-1=2py^2

设p是6k+1型的奇素数,运用Pell方程px2-3y2=1的最小解、同余式、平方剩余、勒让德符号的性质等初等方法证明了当p=3n(n+1)+1≡1,7(mod8)(n为单数)为奇素数,且2n+1为奇素数时,指数Diophantine方程x3-1=2py2无正整数解.

关于Diophantine方程x^3+1=py^2

在素数p=3(8t+4)(8t+5)+1和p=3(8t+3)(8t+4)+1的情形下,运用初等数论的方法给出了丢番图方程x3+1=py2无正整数解的充分条件,并得到无数个6k+1型的素数p使得方程x3+1=py2无正整数解.

关于Diophantine方程x^3+1=3py^2

设p是奇素数,研究丢番图方程x3+1=3py2正整数解的情况.利用初等数论的方法得到了丢番图方程x3+1=3py2无正整数解的若干充分条件.

关于指数Diophantine方程x^3+1=Dy^2

设D是6k+1型的奇素数,运用Pell方程px2-3y2=1的最小解、同余式、平方剩余等初等方法给出了:当D=12t2+1(t是奇数)时,Diophantine方程x3+1=Dy2无正整数解的一个充分条件.

关于Diophantine方程x^3±8=3Dy^2

利用初等数论的方法证明了:如果D是适合D≡5(mod8)的奇素数,则方程x3+8=3Dy2无正整数解;如果D是适合D≡7(mod8)的奇素数,则方程x3-8=3Dy2无正整数解。

关于Diophantine方程x^3±1=Dy^2

利用数论中的同余,勒让德符号的性质及其它一些方法,研究丢番图方程x3±1=Dy2(D=D1p,D是无平方因子的正整数,其中D1是不能被3或6k+1之形的素数整除的正整数,p是奇素数,p=3(24r+19)(24r+20)+1,r是正整数)的解的情况.证明了当D1≡7(mod12)时,方程x3+1=Dy2无正整数解;当D1≡5,14,17,23(mod24)时,方程x3-1=Dy2无正整数解.推进了该类三次丢番图方程的研究.

关于Diophantine方程x^3±1=2py^2

设p是奇素数,证明了当p=6(4s+1)+1,其中s是非负整数时,方程x3-1=2py2仅有整数解(x,y)=(1,0);当p=6(4s+2)+1,其中s是非负整数时,方程x3+1=2py2仅有整数解(x,y)=(-1,0).

关于Diophantine方程x^3-1=py^2

利用初等数论的方法得到丢番图方程x3-1=py2无正整数解的一个充分条件。设p是奇素数,证明了当p=3(4k+3)(4k+4)+1,其中k是非负整数,则方程x3-1=py2无正整数解。

Diophantine方程x^3+8=py^2有本原正整数解的必要条件

设p是奇素数.运用Pell方程的性质证明了:如果方程x^3+8=py^2有适合gcd(x,y)=1的正整数解(x,y),则必有p≡1,7(mod 24).