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.

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 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...

For $1\le c\le p-1$, let $E_1,E_2,\cdots ,E_m$ be fixed numbers of the set $\{0,1\}$, and let $a_1,a_2,\cdots ,a_m$ $(1\le a_i\le p$, $i=1,2,\cdots ,m)$ be of opposite parity with $E_1,E_2,\cdots ,E_m$ respectively such that $a_1{a}_2\cdots a_m\equiv c(modp)$. Let $$N(c,m,p)=\frac{1}{2^{m-1}}\underset{a_1{a}_2\cdots a_m\equiv c(modp)}{\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_m=1}^{p-1}}(1-{(-1)}^{a_1+E_1})(1-{(-1)}^{a_2+E_2})\cdots (1-{(-1)}^{a_m+E_m}).$$ We are interested in the mean value of the sums $$\sum _{c=1}^{p-1}{E}^2(c,m,p),$$ where $E(c,m,p)=N(c,m,p)-\left({(p-1)}^{m-1}\right)/\left(2^{m-1}\right)$ for the odd prime $p$ and any integers $m\ge 2$. When $m=2$, $c=1$, it is the Lehmer problem. In this paper, we generalize the Lehmer problem and use analytic method to give an interesting asymptotic formula of the generalized Lehmer problem.

For the general modulo $q\ge 3$ and a general multiplicative character $\chi$ modulo $q$, the upper bound estimate of $\left|S\right(m,n,1,\chi ,q\left)\right|$ is a very complex and difficult problem. In most cases, the Weil type bound for $\left|S\right(m,n,1,\chi ,q\left)\right|$ is valid, but there are some counterexamples. Although the value distribution of $\left|S\right(m,n,1,\chi ,q\left)\right|$ is very complicated, it also exhibits many good distribution properties in some number theory problems. The main purpose of this paper is using the estimate for $k$-th Kloosterman sums and analytic method to study the asymptotic...

Let $a$, $b$, $c$, $r$ be positive integers such that $a^2+b^2=c^r$, $min(a,b,c,r)>1$, $gcd(a,b)=1,a$ is even and $r$ is odd. In this paper we prove that if $b\equiv 3\left(mod4\right)$ and either $b$ or $c$ is an odd prime power, then the equation $x^2+b^y=c^z$ has only the positive integer solution $(x,y,z)=(a,2,r)$ with $min(y,z)>1$.