相关论文: On the Diophantine equation X^{2N} + 2^{2 alpha} 5…

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 $gcd(x,y)=1$.

We prove that the Diophantine equation N X^2 + 2^L 3^M = Y^N has no solutions (N,X,Y,L,M) in positive integers with N > 1 and gcd(NX,Y) = 1, generalizing results of Luca, Wang and Wang, and Luca and Soydan. Our proofs use results of Bilu,…

In this paper, we give all the solutions of the Diophantine equation x^2+7^{alpha}.11^{beta}=y^n, in nonnegative integers x, y, n>=3 with x and y coprime, except for the case when alpha.x is odd and beta is even.

In this work, we prove the following result(Theorem 1): Suppose that n is a positive integer, p an odd prime, and such that either n is congruent to 0 modulo4 and p congruent to 3 modulo8; or alternatively, n is congruent to 2 modulo4 and p…

Let $p$ be a prime number with $p>3$, $p\equiv 3\pmod{4}$ and let $n$ be a positive integer. In this paper, we prove that the Diophantine equation $(5pn^{2}-1)^{x}+(p(p-5)n^{2}+1)^{y}=(pn)^{z}$ has only the positive integer solution…

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…

Let $n$ be a positive integer and consider the Diophantine equation of generalized Fermat type $x^2+y^{2n}=z^3$ in nonzero coprime integer unknowns $x,y,z$. Using methods of modular forms and Galois representations for approaching…

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 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 consider the Diophantine equation $(-1)^{\alpha}p^x+(-1)^{\beta}(2^k(2p+1))^y=z^2$ for Sophie Germain prime $p$ with $\alpha, \beta \in\{0,1\}$, $\alpha\beta=0$ and $k\geq 0$. First, for $p=2$, we solve three Diophantine…

Let $n$ be a positive integer. We discuss pairs of distinct odd primes $p$ and $q$ not dividing $n$ for which the Diophantine equations $pq=x^2+ny^2$ have integer solutions in $x$ and $y$. As its examples we classify all such pairs of $p$…

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…

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 show that the Diophantine equation given by X^3+ XYZ = Y^2+Z^2+5 has no integral solution. As a consequence, we show that the family of elliptic curve given by the Weierstrass equations Y^2-kXY = X^3 - (k^2+5) has no integral point.

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…

In this paper we find non-negative integer solutions for exponential Diophantine equations of the type $p \cdot 3^x+ p^y=z^2,$ where $p$ is a prime number. We prove that such equation has a unique solution…

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.

In an earlier paper, Tatong and Suvarnamani explores the Diophantine equation $p^x + p^y = z^2$ for a prime number $p$. In that paper they find some solutions to the equation for $p=2, 3$. In this paper, we look at a general version of this…

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 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.