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.     

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...     

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).     

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.     

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.     

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.     

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 equationxyz=x+y+z=1 in integers of a cubic field

We determine all cubic number fields such that the title equation has a solution in the ring of integers of the field.I am grateful to H. M. Edgar for suggesting this problem.     

A note on the simultaneous Pell equations $$x^2-(a^2-1)y^2=1$$ x 2 - ( a 2 - 1 ) y 2 = 1 and $$y^2-bz^2=1$$ y 2 - b z 2 = 1

Periodica Mathematica Hungarica - Let $$a>1,b$$ be two positive integers where the square-free part of b is 2pq with p, q two distinct odd primes. Recently, Cipu (Proc Am Math Soc...     

On the diophantine equationD 1 x 2+D 2 m =4y n

For anyD ₁,D ₂, leth(-D ₁ D ₂) denote the class number of the imaginary quadratic field . In this paper we prove that the equationD ₁ x ²+D ₂ ^m =4y ⁿ.D ₁,D ₂,x, y, m, n, gcd (D ₁x,D ₂y=1,2m,n an odd prime,nh(-D ₁ D ₂, has only a finite number of solutions (D ₁,D ₂,x,y,m,n) withn>5. Moreover, the solutions satisfy 4y ⁿ相似文献   

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     

On the Diophantine equation $$Cx^{2}+D=2y^{q}$$ C x 2 + D = 2 y q

The Ramanujan Journal - Let C and D denote positive integers such that $$CD>1$$ . In this paper we investigate the solvability of the Diophantine equation $$Cx^{2}+D=2y^{q}$$ , in positive...