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     

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.     

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).     

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...     

关于不定方程x~2-kxy+y~2+px=0

设p为奇素数.讨论了不定方程x~2-kxy+y~2+px=0,给出了这类方程求解的一些必要条件.     

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.     

关于不定方程x3-1=749y2

利用同余式、平方剩余、Pell方程的解的性质、递归序列证明了:不定方程x³-1=749y²仅有整数解(x,y)=(1,0).     

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.     

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).     

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 $y \in \mathbb{D}$ .     

A Conjecture Concerning the Pure Exponential Diophantine Equation a^x ＋ b^y = c^z

Let a, b, c, r be fixed positive integers such that a^2 ＋ b^2 = c^r, min（a, b, c, r） 〉 1 and 2 r. In this paper we prove that if a ≡ 2 （mod 4）, b ≡ 3 （mod 4）, c 〉 3.10^37 and r 〉 7200, then the equation a^x ＋ b^y = c^z only has the solution （x, y, z） = （2, 2, r）.     

Über die Funktionalgleichungf(a 0+a 1 x+a 2 y+a 3 xy)+g(b 0+b 1 x+b 2 y+b 3 xy=h(x)+k(y)

Ohne ZusammenfassungHerrn ProfessorG. Alexits zum 70. Geburtstag gewidmet     

A ternary Diophantine inequality by primes with one of the form $$p=x^2+y^2+1$$ p = x 2 + y 2 + 1

The Ramanujan Journal - In this paper we solve the ternary Piatetski-Shapiro inequality with prime numbers of a special form. More precisely we show that, for any fixed $$1<\frac{427}{400}$$...