The diophantine equationx 2+2 k =y n

The author wishes to thank the referee most sincerely for the observation immediately following the statement of the theorem.     

A note on the diophantine equationx 2+ 4D=y p

LetD be a positive square free integer, and leth(–D) denote the class number of . Furthermore letp be an odd prime with . In this note we prove that ifp {5, 7} orp>3·10⁶, then the equation , has no positive integer solution (x, y).     

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   

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 $y \in \mathbb{D}$ .     

On the diophantine equation y2 = 4qn + 4q + 1

It is known that a certain class of [n, k] codes over GF(q) is related to the diophantine equation y² = 4qⁿ + 4q + 1 (1). In Parts I and II of this paper, two different, and in a certain sense complementary, methods of approach to (1) are discussed and some results concerning (1) are given as applications. A typical result is that the only solutions to (1) are (y, n) = (5, 1), (7, 2), (11, 3) when q = 3 and (y, n) = (2q + 1, 2) when q = 3^f, f >- 2.     

Integral solutions in arithmetic progression for y2=x3+k

Integral solutions toy ²=x ³+k, where either thex's or they's, or both, are in arithmetic progression are studied. When both thex's and they's are in arithmetic progression, then this situation is completely solved. One set of solutions where they's formed an arithmetic progression of length 4 had already been constructed. In this paper, we construct infinitely many sets of solutions where there are 4x's in arithmetic progression and we disprove Mohanty's Conjecture [8] by constructing infinitely many sets of solutions where there are 4, 5 and 6y's in arithmetic progression.     

On the equationx(x +d 1)…(x + (k - 1)d 1) =y(y +d 2)…(y + (mk - 1)d 2)

For given positive integersm ≥ 2,d ₁ andd ₂, we consider the equation of the title in positive integersx, y andk ≥ 2. We show that the equation implies thatk is bounded. For a fixedk, we give conditions under which the equation implies that max(x, y) is bounded. Dedicated to the memory of Professor K G Ramanathan     

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 Fn+Fm=2a

In this paper, we find all the solutions of the title Diophantine equation in positive integer variables (n, m, a), where F_k is the kth term of the Fibonacci sequence. The proof of our main theorem uses lower bounds for linear forms in logarithms (Baker's theory) and a version of the Baker-Davenport reduction method in diophantine approximation.     

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).     

On the equationx t=x t+q in categories

We will show that a categoryC divides a monoid that satisfies the equationx ^t=x ^t+q iff each monoid inC satisfies the same equation.