The al-husayn equation x 4 + y 2 = z 2

We study the set of all natural solutions of the equation x ⁴ + y ² = z ², obtain general formulas describing all such solutions, and prove their equivalence.     

关于不定方程x~2+4~n=y~(13)(n=4,5,6)的整数解

在高斯整环中,利用代数数论与同余理论的方法,讨论了不定方程x~2+4~n=y~(13)(n=4,5,6)的整数解问题,得出了当n=4,5时无整数解;n=6是仅有整数解(x,y)=(64,2)和(x,y)=(-64,2)的结论,推进了不定方程整数解的研究.     

On the Diophantine equation x 2+5 m =y n

In this paper we consider the Diophantine equation x ²+5 ^m =y ⁿ , n>2, m>0. We prove that the equation has no positive integer solutions when 2∤ m, nor when 2∣m under the additional condition (x,y)=1, with the help of Bilu, Hanrot, and Voutier’s deep result in (J. Reine Angew. Math. 539:75–122, 2001). Supported by the 973 Grant of P.R.C and SRFDP 20040284018.     

The exponential diophantine equation x y + y x = y 2 with xy odd

For any fixed positive integer D which is not a square, let (u, υ) = (u ₁, υ ₁) be the fundamental solution of the Pell equation u ² ? Dυ ² = 1. Further let $\mathbb{D}$ be the set of all positive integers D such that D is odd, D is not a square and gcd(D, υ ₁) > max(1, √D/8). In this paper we prove that if (x, y, z) is a positive integer solution of the equation x ^y + y ^x = z ² satisfying gcd(x, y) = 1 and xy is odd, then either $x \in \mathbb{D}$ or $y \in \mathbb{D}$ .     

The equation $$x^4+2^ny^4=z^4$$ x 4 + 2 n y 4 = z 4 in algebraic number fields

Acta Mathematica Hungarica - Let n be a positive integer. We show that if the equation $$(1) \qquad \qquad \qquad x^4+2^ny^4=z^4$$ has a solution (x,y,z) in a cubic number field K with $$xyz \neq...     

关于丢番图方程（195n）x+（28n）y=（197n）z

运用同余及元素阶的性质,证明了对任意的正整数n,丢番图方程(195n)x+(28n)y=(197n)z仅有正整数解(x, y, z)=(2,2,2)。     

关于丢番图方程x￣2+D=y￣n

本文运用Ｂａｋｅｒ方法证明了：当Ｄ＝６７时，方程ｘ２＋Ｄ＝ｙｎ，ｘ，ｙ，ｎ∈Ｎ，ｎ＞２，仅有解（ｘ，ｙ，ｎ）＝（１１０，２３，３）；当Ｄ＝４３或１６３时，该方程无解     

关于丢番图方程x￣2+D=y￣n

本文运用Ｂａｋｅｒ方法证明了：当Ｄ＝６７时，方程ｘ２＋Ｄ＝ｙｎ，ｘ，ｙ，ｎ∈Ｎ，ｎ＞２，仅有解（ｘ，ｙ，ｎ）＝（１１０，２３，３）；当Ｄ＝４３或１６３时，该方程无解     

关于Diophantine方程x3+1=py2

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.     

On the Diophantine equation $$x^2+3^a41^b=y^n $$ x 2 + 3 a 41 b = y n

Periodica Mathematica Hungarica - In this paper we find all positive integer solutions (x, y, n, a, b) of the equation in the title for non negative integers a...     

关于Diophantine方程(n+1)+(n+2)y=nz

设n是正整数.本文证明了:方程(n+1)+(n+2)y=nz仅当n=3时有正整数解(y,z)=(1,2).     

关于Diophantine方程（44n）x＋（117n）y=（125n）z 的整数解

在Jeismanowicz猜想的基础上,利用初等方法证明了对任意的正整数n, Diophantine方程（44n）x＋（117n）y=（125n）z 仅有正整数解（x, y, z）=（2,2,2）。     

A note on the diophantine equation x 2 + b y = c z

Let a, b, c, r be positive integers such that a ² + b ² = c ^r , min(a, b, c, r) > 1, gcd(a, b) = 1, a is even and r is odd. In this paper we prove that if b ≡ 3 (mod 4) and either b or c is an odd prime power, then the equation x ² + b ^y = c ^z has only the positive integer solution (x, y, z) = (a, 2, r) with min(y, z) > 1.     

Il sistema diofanteo N+1=x2, 3N+1=y2, 8N+1=z2

Sunto Nel 1969 A. Baker e H. Davenport hanno dimostrato che il sistema3x ² −2=y ² ,8x ² −7=z ² ha soltanto le soluzioni |x|=|y|=|z|=1; |x|=11, |y|=19, |z|=31. Una seconda dimostrazione di P. Kanagasabapathy — Tharmambikai Ponmudurai è apparsa nell'ottobre 1975. L'A. dà una terza dimostrazione che consente, con l'uso dei metodi classici dell'aritmetica, di controllare i ragionamenti in ogni punto. A Dario Graffi nel suo 70° compleanno Entrata in Redazione il 26 novembre 1975.