Let $k$ be a finite extension of ${\mathbb{Q}}_p$, let $k_1$, respectively $k_3$, be the division fields of level $1$, respectively $3$, arising from a Lubin-Tate formal group over $k$, and let $\Gamma =$ Gal($k_3/k_1$). It is known that the valuation ring $k_3$ cannot be free over its associated order $𝔄$ in $K\Gamma$ unless $k={\mathbb{Q}}_p$. We determine explicitly under the hypothesis that the absolute ramification index of $k$ is sufficiently large.

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.

A sequence is called $k$-automatic if the $n$’th term in the sequence can be generated by a finite state machine, reading $n$ in base $k$ as input. We show that for many multiplicative functions, the sequence ${\left(f\left(n\right)\text{mod}v\right)}_{n\ge 1}$ is not $k$-automatic. Among these multiplicative functions are ${\gamma }_m\left(n\right),{\sigma }_m\left(n\right),\mu \left(n\right)$ et $\phi \left(n\right)$.

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.

Let $n$ be a positive integer, and $C_n\left(r\right)$ the set of all $n\times n$ $r$-circulant matrices over the Boolean algebra $B=\{0,1\}$, $G_n={\bigcup }_{r=0}^{n-1}{C}_n\left(r\right)$. For any fixed $r$-circulant matrix $C$ ($C\ne 0$) in $G_n$, we define an operation “$*$” in $G_n$ as follows: $A*B=ACB$ for any $A,B$ in $G_n$, where $ACB$ is the usual product of Boolean matrices. Then $(G_n,*)$ is a semigroup. We denote this semigroup by $G_n\left(C\right)$ and call it the sandwich semigroup of generalized circulant Boolean matrices with sandwich matrix $C$. Let $F$ be an idempotent element in $G_n\left(C\right)$ and $M\left(F\right)$ the maximal subgroup...

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 $min(y,z)>1$.