Let a,b and c be relatively prime positive integers such that a²+b² = c². We prove that if $b\equiv 0\left(mod{2}^r\right)$ and $b\equiv \pm 2^r\left(moda\right)$ for some non-negative integer r, then the Diophantine equation $a^x+b^y=c^z$ has only the positive solution (x,y,z) = (2,2,2). We also show that the same holds if c ≡ -1 (mod a).

This paper investigates the system of equations $$x^2+a{y}^m=z_1^2,x^2-a{y}^m=z_2^2$$ in positive integers $x$, $y$, $z_1$, $z_2$, where $a$ and $m$ are positive integers with $m\ge 3$. In case of $m=2$ we would obtain the classical problem of congruent numbers. We provide a procedure to solve the simultaneous equations above for a class of the coefficient $a$ with the condition $gcd(x,z_1)=gcd(x,z_2)=gcd(z_1,z_2)=1$. Further, under same condition, we even prove a finiteness theorem for arbitrary nonzero $a$.

Let ℤ be the set of integers, and let (m,n) be the greatest common divisor of the integers m and n. Let p ≡ 1 (mod 4) be a prime, q ∈ ℤ, 2 ∤ q and p=c²+d²=x²+qy² with c,d,x,y ∈ ℤ and c ≡ 1 (mod 4). Suppose that (c,x+d)=1 or (d,x+c) is a power of 2. In this paper, by using the quartic reciprocity law, we determine $q^{[p/8]}\left(modp\right)$ in terms of c,d,x and y, where [·] is the greatest integer function. Hence we partially solve some conjectures posed in our previous two papers.

Let $\mathbb{Z}$, $\mathbb{N}$ be the sets of all integers and positive integers, respectively. Let $p$ be a fixed odd prime. Recently, there have been many papers concerned with solutions $(x,y,n,a,b)$ of the equation $x^2+2^a{p}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$, $b\ge 0.$ And all solutions of it have been determined for the cases $p=3$, $p=5$, $p=11$ and $p=13$. In this paper, we mainly concentrate on the case $p=3$, and using certain recent results on exponential diophantine equations including the famous Catalan equation, all solutions $(x,y,n,a,b)$ of the equation $x^2+2^a·{17}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$,...

We show that any factorization of any composite Fermat number $F_m={2^2}^m+1$ into two nontrivial factors can be expressed in the form $F_m=(k{2}^n+1)(\ell 2^n+1)$ for some odd $k$ and $\ell ,k\ge 3,\ell \ge 3$, and integer $n\ge m+2,3n<2^m$. We prove that the greatest common divisor of $k$ and $\ell$ is 1, $k+\ell \equiv 0mod{2}^n,max(k,\ell )\ge F_{m-2}$, and either $3|k$ or $3|\ell$, i.e., $3h{2}^{m+2}+1|F_m$ for an integer $h\ge 1$. Factorizations of $F_m$ into more than two factors are investigated as well. In particular, we prove that if $F_m={(k{2}^n+1)}^2(\ell 2^j+1)$ then $j=n+1,3|\ell$ and $5|\ell$.