This allows us to deduce that for each integer $n\geq 3$, there exist infinitely many squarefree composite numbers $N$ with $n$ prime factors and empty rational Korselt sets.
📌 3 Maddede Bu Makale
🎯
Amaç & Hipotez:Let $N$ be a positive integer, and $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{Q}\setminus \{0,N\}$ with $\gcd(\alpha_{1}, \alpha_{2})=1$. $N$ is called an $\alpha$-Korselt number, equivalently $\alpha$ is said an $N$-Korselt base, if $\alpha_{2}p-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $p$ of $N$.
📊
Temel Bulgular & Veri:The set of $N$-Korselt bases in $\mathbb{Q}$ is denoted by $\mathbb{Q}$-$\mathcal{KS}(N)$ and called the set of rational Korselt bases of $N$.In this paper rational Korselt bases are deeply studied, where we give in details their belonging sets and their forms in some cases.
💡
Sonuç & Katkı:This allows us to deduce that for each integer $n\geq 3$, there exist infinitely many squarefree composite numbers $N$ with $n$ prime factors and empty rational Korselt sets.
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:prime number · squarefree composite number · Korselt number
Abstract
Let $N$ be a positive integer, and $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{Q}\setminus \{0,N\}$ with $\gcd(\alpha_{1}, \alpha_{2})=1$. $N$ is called an $\alpha$-Korselt number, equivalently $\alpha$ is said an $N$-Korselt base, if $\alpha_{2}p-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $p$ of $N$. The set of $N$-Korselt bases in $\mathbb{Q}$ is denoted by $\mathbb{Q}$-$\mathcal{KS}(N)$ and called the set of rational Korselt bases of $N$.In this paper rational Korselt bases are deeply studied, where we give in details their belonging sets and their forms in some cases. This allows us to deduce that for each integer $n\geq 3$, there exist infinitely many squarefree composite numbers $N$ with $n$ prime factors and empty rational Korselt sets.