Related papers: Congruences for $q^{[p/8]}\pmod p$ II

Let $\Bbb Z$ be the set of integers, and let $(m,n)$ be the greatest common divisor of integers $m$ and $n$. Let $p\equiv 1\mod 4$ be a prime, $q\in\Bbb Z$, $2\nmid q$ and $p=c^2+d^2=x^2+qy^2$ with $c,d,x,y\in\Bbb Z$ and $c\e 1\mod 4$.…

For a prime $p\equiv 3$ $(\text{mod }4)$, let $h(-8p)$ and $h(8p)$ be the class numbers of $\mathbb{Q}(\sqrt{-2p})$ and $\mathbb{Q}(\sqrt{2p})$, respectively. Let $\Psi(\xi)$ be the Hirzebruch sum of a quadratic irrational $\xi$. We show…

Suppose that $p$ is an odd prime and $m$ is an integer not divisible by $p$. Sun and Tauraso [Adv. in Appl. Math., 45(2010), 125--148] gave $\sum_{k=0}^{n-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=0}^{n-1}\binom{2k}{k+d}/(km^k)$ modulo $p$ for…

Let $p$ be an odd prime and let $m\not\equiv 0\pmod p$ be a rational p-adic integer. In this paper we reveal the connection between quartic residues and the sum $\sum_{k=0}^{[p/4]}\binom{4k}{2k}\frac 1{m^k}$, where $[x]$ is the greatest…

Nagell proved that for each prime $p\equiv 1\pmod{3}$, $p > 7$, there is a prime $q<2p^{1/2}$ that is a cubic residue modulo $p$. Here we show that for each fixed $\epsilon > 0$, and each prime $p\equiv 1\pmod{3}$ with $p > p_0(\epsilon)$,…

We show that substantially more than a quarter of the odd integers of the form $pq$ up to $x$, with $p,q$ both prime, satisfy $p\equiv q\equiv 3 \pmod 4$.

Let $q$ be a prime. We classify the odd primes $p\neq q$ such that the equation $x^2\equiv q\pmod{p}$ has a solution, concretely, we find a subgroup $\mathbb{L}_{4q}$ of the multiplicative group $\mathbb{U}_{4q}$ of integers relatively…

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.

In this paper, we mainly prove two congruence conjecture of Z.-W. Sun. Let $p\equiv3\pmod 4$ be a prime. Then $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{8^k}\equiv-\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{(-16)^k}\pmod{p^3}.$$ And for any odd prime…

Given a squarefree positive integer $d$, we want to find integers (or rational numbers with denominators not divisible by large primes) $a_0,a_1,a_2,\ldots$ such that for sufficiently large primes $p$ we have $\sum_{k=0}^{p-1}a_k\equiv…

Let $p\equiv3\pmod{4}$ be a prime and $r$ a positive integer. We show that $$ \prod_{k=1}^{(p^{2r}-1)/2}\frac{4k-1}{4k+1}\equiv1\pmod{p^2}. $$ This confirms a recent conjecture of Guo.

Let $p$ be a prime greater than 3. In the paper we mainly determine $\sum_{k=0}^{[p/4]}\binom{4k}{2k}(-1)^k$, $\sum_{k=0}^{[p/3]}\binom{3k}k, \sum_{k=0}^{[p/3]}\binom{3k}k(-1)^k$ and $\sum_{k=0}^{[p/3]}\binom{3k}k(-3)^k$ modulo $p$, where…

Let $\overline{p}_{k}(n)$ denote the number of overpartition $k$-tuples of $n$. In 2023, Saikia \cite{saikia} conjectured the following congruences: \begin{align*} \overline{p}_{q}(8n+2)& \equiv 0 \pmod{4},\quad \overline{p}_{q}(8n+3)\equiv…

Let $p$ be an odd prime. In this paper we investigate quadratic residues modulo $p$ and related permutations, congruences and identities. If $a_1<\ldots<a_{(p-1)/2}$ are all the quadratic residues modulo $p$ among $1,\ldots,p-1$, then the…

Let $p^k m^2$ be an odd perfect number with special prime $p$. In this article, we provide an alternative proof for the biconditional that $\sigma(m^2) \equiv 1 \pmod 4$ holds if and only if $p \equiv k \pmod 8$. We then give an application…

In this article, we consider primes $p \equiv 5 \pmod 8$ and are able to prove that $p \equiv 5 \pmod {16}$ if $2p$ is a congruent number.

Let $p$ be a fixed odd prime and $Q(x,y,z)=ax^2+bxy+cy^2+dxz+eyz+fz^2$ be a fixed quadratic form in $\mathbb{Z}[x,y,z]$ which is non-degenerate in $\mathbb{F}_p[x,y,z]$ and $(a(4ac-b^2),p)=1.$ Let $(x_0,y_0,z_0)$ be a fixed point in…

Let $p$ be a prime greater than $3$ and let $a$ be a rational p-adic integer. In this paper we try to determine $\sum_{k=1}^{[p/3]}\binom{3k}ka^k\pmod p$, and real the connection between cubic congruences and the sum…

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

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