相关论文: On the Diophantine equation x^2+7^{alpha}.11^{beta…

The objective of the paper is to determine the complete solutions for the Diophantine equation $x^2 + 3^{\alpha}113^{\beta} = y^{\mathfrak{n}}$ in positive integers $x$ and $y$ (where $x, y \geq 1$), non-negative exponents $\alpha$ and…

In this note, we find all the solutions of the Diophantine equation x^2 +2^a.3^b.11^c=y^n in nonnegative integers a, b, c, x, y, n>= 3 with x and y coprime.

The object of this paper is to give a new proof of all the solutions of the Diophantine equation x^2+11^m=y^n; in positive integers x, y with odd m>1 and n>=3.

We prove that for each odd prime p, positive integer alpha, and non-negative integers beta and gamma, the Diophantine equation X^{2N} + 2^{2 alpha} 5^{2 beta} p^{2 gamma} = Z^5 has no solution with X, Z, N in Z^+, N > 1, and gcd(X,Z) = 1.

We find all solutions of Diophantine equation x^{2}+11^{2k} = y^{n} where x>=1, y>=1, n>=3 and k is natural number. We give p-adic interpretation of this equation.

We give the complete solution in integers $(n,a,b,x,y)$ of the title equation when $\gcd(x,y)=1$, except for the case when $xab$ is odd.

Suppose that $x$ is odd, $n\geq7$ and $p\notin\{2,5\}$ are primes. In this paper, we prove that the Diophantine equations $x^{2}\pm5^{\alpha}p^{n}=y^{n}$ have no solutions in positive integers $\alpha,x,y$ with $gcd(x,y)=1$.

In this paper, we gave solutions of the Diophantine equations 16^{x}+p^{y}=z^{2}, 64^{x}+p^{y}=z^{2} where p is an odd prime, n is a positive integer and x,y,z are non-negative integers. Finally we gave a generalization of the Diophantine…

In this note we consider the title Diophantine equation from both theoretical as well as experimental point of view. In particular, we prove that for $k=4, 6$ and each choice of the signs our equation has infinitely many co-prime positive…

It is a generalization of Pell's equation $x^2-Dy^2=0$. Here, we show that: if our Diophantine equation has a particular integer solution and $ab$ is not a perfect square, then the equation has an infinite number of solutions; in this case…

We consider the Diophantine equation $7x^{2} + y^{2n} = 4z^{3}$. We determine all solutions to this equation for $n = 2, 3, 4$ and $5$. We formulate a Kraus type criterion for showing that the Diophantine equation $7x^{2} + y^{2p} = 4z^{3}$…

We solve the Diophantine equation $Y^2=X^3+k$ for all nonzero integers $k$ with $|k| \leq 10^7$. Our approach uses a classical connection between these equations and cubic Thue equations. The latter can be treated algorithmically via lower…

We investigate the solvability of the Diophantine equation in the title, where $d>1$ is a square-free integer, $p, q$ are distinct odd primes and $x,y,a,b$ are unknown positive integers with $\gcd(x,y)=1$. We describe all the integer…

We prove that for given integers b and c, the diophantine equation x^2+bx+c=y^2, has finitely many integer solutions(i.e. pairs in ZxZ),in fact an even number of such solutions(including the zero or no solutions case).We also offer an…

In this study we find all solutions of the Diophantine equation $B_{n_{1}}+B_{n_{2}}=2^{a_{1}}+2^{a_{2}}+2^{a_{3}}$ in positive integer variables $(n_{1},n_{2},a_{1},a_{2},a_{3}),$ where $B_{n}$ denotes the $n$-th balancing number.

In this work, we accomplish three goals. First, we determine the entire family of positive integer solutions to the three- variable Diophantine equation, xy=z^2; for n=2,3,4,5,6. For n=2, we obtain a 3-parameter family of solutions; for…

Using elementary number theory we study Diophantine equations over the rational integers of the following form, $y^2=(x+a)(x+a+k)(x+b)(x+b+k)$, $y^2=c^2x^4+ax^2+b$ and $y^2=(x^2-1)(x^2-\alpha^2)(x^2-(\alpha+1)^2).$ We express their integer…

We prove that the Diophantine equation (a^2 c x^k - 1)(b^2 c y^k - 1) = (abc z^k - 1)^2 has no solutions in positive integers with x, y, z > 1, k \geq 7 and a^2x^k \neq b^2y^k.

Given odd, coprime integers $a$, $b$ ($a>0$), we consider the Diophantine equation $ax^2+b^{2l}=4y^n$, $x, y\in\Bbb Z$, $l \in \Bbb N$, $n$ odd prime, $\gcd(x,y)=1$. We completely solve the above Diophantine equation for…

First, we consider the equation $ax^2 - by^2 + c = 0$, with $a,b \in N*$ and $c \in Z*$, which is a generalization of Pell's equation. Here, we show that: if this equation has an integer solution and $ab$ is not a perfect square, then it…