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 $5|\ell$.

Let ℤ be the set of integers, and let (m,n) be the greatest common divisor of integers m and n. Let p be a prime of the form 4k+1 and p = c²+d² with c,d ∈ ℤ, $d=2^{r}d₀$ and c ≡ d₀ ≡ 1 (mod 4). In the paper we determine ${(b+\surd (b²+4^{\alpha })/2)}^{(p-1)/4)}\left(modp\right)$ for p = x²+(b²+4α)y² (b,x,y ∈ ℤ, 2∤b), and ${(2a+\surd 4a²+1)}^{(p-1)/4}\left(modp\right)$ for p = x²+(4a²+1)y² (a,x,y∈ℤ) on the condition that (c,x+d) = 1 or (d₀,x+c) = 1. As applications we obtain the congruence for $U_{(p-1)/4}\left(modp\right)$ and the criterion for $p|U_{(p-1)/8}$ (if p ≡ 1 (mod 8)), where Uₙ is the Lucas sequence given by U₀ = 0, U₁ = 1 and...

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(h{p}^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{k)/m^k\left(modp²\right)}}$, where h and m are p-adic integers with m ≢ 0 (mod p). For example, we show that if h ≢ 0 (mod p) and $p^a>3$, then $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(h{p}^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{{k)(-h/2)}^k\equiv ((1-2h)/\left(p^a\right))(1+h({(4-2/h)}^{p-1}-1))\left(modp²\right)}}$, where (·/·) denotes the Jacobi symbol. Here is another remarkable congruence: If $p^a>3$ then $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(p^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{{k)(-1)}^k\equiv 3^{p-1}(p^a/3)\left(modp²\right)}}$.

Let E be an interval in the unit interval [0,1). For each x ∈ [0,1) define dₙ(x) ∈ 0,1 by $dₙ\left(x\right):={\sum }_{i=1}^n{1}_E\left(2^{i-1}x\right)\left(mod2\right)$, where t is the fractional part of t. Then x is called a normal number mod 2 with respect to E if $N^{-1}{\sum }_{n=1}^{N}dₙ\left(x\right)$ converges to 1/2. It is shown that for any interval E ≠(1/6, 5/6) a.e. x is a normal number mod 2 with respect to E. For E = (1/6, 5/6) it is proved that $N^{-1}{\sum }_{n=1}^{N}dₙ\left(x\right)$ converges a.e. and the limit equals 1/3 or 2/3 depending on x.