相关论文: There are no socialist primes less than 10^9

Paul Erdos posed the following question: Is there a prime number $p>5$ such that the residues of $2!$, $3!$,\ldots, $(p-1)!$ modulo $p$ all are distinct? In this short note, we prove that there are no such prime numbers.

We investigate the existence of primes $p > 5$ for which the residues of $2!$, $3!$, \dots, $(p-1)!$ modulo $p$ are all distinct. We describe the connection between this problem and Kurepa's left factorial function, and report that there…

We present improved algorithms for computing the left factorial residues $!p=0!+1!+...+(p-1)! \!\mod p$. We use these algorithms for the calculation of the residues $!p\!\mod p$, for all primes $p$ up to $2^{40}$. Our results confirm that…

Paul Erdos posed the following question: Is there a prime number $p>5$ such that the residues of $2!$, $3!$,\ldots, $(p-1)!$ modulo $p$ all are distinct. In this short note, we give the negative answer on this question in an elementary way.

Kurepa's conjecture states that there is no odd prime $p$ that divides $!p=0!+1!+\cdots+(p-1)!$. We search for a counterexample to this conjecture for all $p<2^{34}$. We introduce new optimization techniques and perform the computation…

Let $p$ be a prime. We define $S(p)$ the smallest number $k$ such that every positive integer is a sum of at most $k$ squares of integers that are not divisible by $p$. In this article, we prove that $S(2)=10$, $S(3)=6$, $S(5)=5$, and…

We prove lower bounds for the number of primes $p \leq N + b$ such that $p-b$ is divisible by $2^{k(N)}$ and has at most $k$ odd prime factors ($k \geq 2$), assuming $2^{k(N)} \leq N^\theta$ for some $\theta > 0$ depending on $k$. The proof…

Let p be any prime, and $p^(\nu_p(n!))$ the maximal power of $p$ dividing $n!$. It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ and all primes…

A prime $p$ is a Wolstenholme prime if $\binom{2p}{p}\equiv2$ mod $p^4$, or, equivalently, if $p$ divides the numerator of the Bernoulli number $B_{p-3}$; a Vandiver prime $p$ is one that divides the Euler number $E_{p-3}$. Only two…

For n=1,2,3,... define S(n) as the smallest integer m>1 such that those 2k(k-1) mod m for k=1,...,n are pairwise distinct; we show that S(n) is the least prime greater than 2n-2 and hence the value set of the function S(n) is exactly the…

We prove that there are no Wieferich's primes $q=2p+1$ where $p \equiv 3 \pmod{4}$ is a prime number

We provide an explicit estimate on the least primitive root mod $p^{2}$. We show, in particular, that every prime $p$ has a primitive root mod $p^{2}$ that is less than $p^{0.99}$.

A Wilson prime is a prime p such that (p-1)! = -1 mod p^2. We report on a search for Wilson primes up to 2 * 10^13, and describe several new algorithms that were used in the search. In particular we give the first known algorithm that…

Let $p$ be a prime. If an integer $g$ generates a subgroup of index $t$ in $(\mathbb Z/p\mathbb Z)^*,$ then we say that $g$ is a $t$-near primitive root modulo $p$. We point out the easy result that each primitive residue class contains a…

In this paper, we confirm some congruences conjectured by V.J.W. Guo and M.J. Schlosser recently. For example, we show that for primes $p>3$, $$…

For any prime $p>3$ and $j\in\mathbb{N}$ we construct examples of non-$FSZ_{p^j}$ groups of order $p^{p^j+2j-1}$. In the special case of $j=1$ this yields groups of order $p^{p+1}$, which is the minimum possible order for a non-$FSZ$…

For any $\epsilon>0$, there exists $q_0(\epsilon)$ such for any $q\ge q_0(\epsilon)$ and any invertible residue class $a$ modulo $q$, there exists a natural number that is congruent to $a$ modulo $q$ and that is the product of exactly three…

This note presents an upper bound for the least prime primitive roots $g^*(p)$ modulo $p$, a large prime. The current literature has several estimates of the least prime primitive root $g^*(p)$ modulo a prime $p\geq 2$ such as $g^*(p)\ll…

Let n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum_{0<i_1<...<i_n<p}(i_1/3)(-1)^{i_1}/(i_1...i_n)=0 (mod p).$$

We prove that there exists an absolute constant $c>0$ such that if an arithmetic progression $\cP$ modulo a prime number $p$ does not contain zero and has the cardinality less than $cp$, then it can not be represented as a product of two…