相关论文: Complete solution of the Diophantine Equation $x^{…
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.
In this paper, we use the method of Thue and Siegel, based on explicit Pade approximations to algebraic functions, to completely solve a family of quartic Thue equations. From this result, we can also solve the diophantine equation in the…
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 title equation, where $p>3$ is a prime number $\not\equiv 7 \pmod 8$, $q$ is an odd prime number and $x,y,n$ are positive integers with $x,y$ relatively prime, is studied. When $p\equiv 3\pmod 8$, we prove (Theorem 2.3) that there are…
A Thue-Mahler equation is a Diophantine equation of the form $$F(X,Y) = a\cdot p_1^{z_1}\cdots p_v^{z_v}, \qquad \gcd(X,Y)=1$$ where $F$ be an irreducible homogeneous binary form of degree at least $3$ with integer coefficients, $a$ is a…
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 completely solve the Diophantine equation $x^2+2^k11^\ell19^m=y^n$ in integers $x,y\geq 1;~ k,\ell, m\geq 0~$ and $n\geq 3$ with $\gcd(x,y)=1$, except the case $2\mid k, 2\nmid \ell m$ and $5\mid n$. We use this result to recover some…
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…
This paper investigates the exponential Diophantine equation of the form $a^x+b=c^y$, where $a, b, c$ are given positive integers with $a,c \ge 2$, and $x,y$ are positive integer unknowns. We define this form as a "Type-I transcendental…
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…
Consider the diophantine equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$ in positive integers $x_1\le x_2$, and $y_1\le y_2$. Each side of the equation is a product of two terms of a given binary recurrence, respectively. In this…
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…
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…
In this paper we investigate Diophantine equations of the form $T^2=G(\overline{X}),\; \overline{X}=(X_{1},\ldots,X_{m})$, where $m=3$ or $m=4$ and $G$ is specific homogenous quintic form. First, we prove that if…
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 solve completely the Lebesgue-Nagell equation x^2+D=y^n, in integers x, y, n>2, for D in the range 1 =< D =< 100.
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.
In this semi-expository article we solve the diophantine equation $m^5+(4 \cdot 5^4 b^4)mn^4 - n^5=1$ for all integers $b \neq 0$. This gives an example of a family of quintic Thue equations that can be solved completely by using nothing…
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…
In this paper, we consider the exponential Diophantine equation $a^{x}+b^{y}=c^{z},$ where $a, b, c$ be relatively prime positive integers such that $a^{2}+b^{2}=c^{r}, r\in Z^{+}, 2\mid r$ with $b$ even. That is $$a=\mid…