Related papers: Quartic residues and sums involving $\binom{4k}{2k…

A variation on the sum-product problem seeks to show that a set which is defined by additive and multiplicative operations will always be large. In this paper, we prove new results of this type. In particular, we show that for any finite…

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n-$th Bernoulli numbers. In this paper, we will generalize…

A formula for the class number $h$ of the imaginary quadratic field $Q(\sqrt{-p}$ is obtained by counting on a specific way the quadratic residues of a prime number of the form $p=4n-1.$ Formulas for the sum of the quadratic residues are…

In this paper, we derive an explicit combinatorial formula for the number of $k$-subset sums of quadratic residues over finite fields.

Suppose that an infinite set $A$ occupies at most $\frac{1}{2}(p+1)$ residue classes modulo $p$, for every sufficiently large prime $p$. The squares, or more generally the integer values of any quadratic, are an example of such a set. By…

Let $p$ be an odd prime, Jianqiang Zhao has established a curious congruence, which is $$ \sum_{i+j+k=p \atop i,j,k > 0} \frac{1}{ijk} \equiv -2B_{p-3}\pmod p , $$ where $B_{n}$ denotes the $n$-th Bernoulli number. In this paper, we will…

Let b be an odd integer such that b=+/-1 (mod 8) and let q be a prime with primitive root 2 such that q does not divide b. We show that if (p(k)) is a sequence of odd primes, with 0<=k<=q-2 such that p(k)=2p(k-1)+b for all 1<=k<=q-2, then…

We establish some supercongruences related to a supercongruence of Van Hamme, such as \begin{align*} \sum_{k=0}^{(p+1)/2} (-1)^k (4k-1)\frac{(-\frac{1}{2})_k^3}{k!^3} &\equiv p(-1)^{(p+1)/2}+p^3(2-E_{p-3})\pmod{p^{4}},\\…

In this paper, we evaluate some series of the form $$\sum_{k=1}^\infty\frac{ak^2+bk+c}{k(3k-1)(3k-2)m^k\binom{4k}k}.$$ For example, we prove that $$\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^{k}}{k(3k-1)(3k-2)\binom{4k}k}=\frac{3}2\pi$$ and…

Let $\mathbb{F}_q$ be a finite field of order $q$, where $q$ is large odd prime power. In this paper, we improve some recent results on the additive energy of the distance set, and on sumsets of the distance set due to Shparlinski (2016).…

We show that if the congruence above holds and $n\mid m$, then the quotient $Q:=m/n$ satisfies $\sum_{p\mid Q} \frac{Q}{p}+1 \equiv 0\pmod{Q}$, where $p$ is prime. The only known solutions of the latter congruence are $Q=1$ and the eight…

We use the representation $T_2(O)$ for $\q(4,q)$ to show that maximal partial ovoids of $\q(4,q)$ of size $q^2-1$, $q=p^h$, $p$ odd prime, $h > 1$, do not exist. Although this was known before, we give a slightly alternative proof, also…

We study the Vapnik-Chervonenkis (VC) dimension of the set of quadratic residues (i.e. squares) in finite fields, $\mathbb F_q$, when considered as a subset of the additive group. We conjecture that as $q \to \infty$, the squares have the…

I present Franz Lemmermeyer's proof that if p is a prime congruent to 9 mod 16 then the class number of Q(p^(1/4)) is congruent to 2 mod 4.

We mainly introduce two new kinds of numbers given by $$R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...)$$ and $$S_n=\sum_{k=0}^n\binom nk^2\binom{2k}k(2k+1)\quad\ (n=0,1,2,...).$$ We find that such numbers have many…

Let $p$ be a prime and ${\mathcal{P}_{p}}$ the set of positive integers which are prime to $p$. We establish the following interesting congruence \[\sum\limits_{\begin{smallmatrix} i+j+k={{p}^{r}} i,j,k\in {\mathcal{P}_{p}}…

In the paper, we establish a new estimate for Kloosterman sum over primes with respect to an arbitrary modulus $q$. This estimate together with some recent results of the second author are applied to the problem of solvability of the…

Motivated by recent works of Sun and Tauraso, we prove some variations on the Green-Krammer identity involving central q-binomial coefficients, such as $$ \sum_{k=0}^{n-1}(-1)^kq^{-{k+1\choose 2}}{2k\brack k}_q \equiv (\frac{n}{5})…

By a very simple argument, we prove that if $l,m,n$ are nonnegative integers then $$\sum_{k=0}^l(-1)^{m-k}\binom{l}{k}\binom{m-k}{n}\binom{2k}{k-2l+m} =\sum_{k=0}^l\binom{l}{k}\binom{2k}{n}\binom{n-l}{m+n-3k-l}. On the basis of this…

Our aim in writing this paper is to answer to both V. E. Hoggatt, JR \cite{hogg} and Wessner\cite{wess} on the next question: find $\sum_{k=0}^n\binom{n}{k}F_{[k]}^p$, for the case $p\equiv 1\, mod\, 4$ and $p\equiv 3\, mod\, 4$. \par The…