Moreover, we prove that for each $\alpha\in \mathbb{Q}\setminus \{0\}$ there exist only finitely many squarefree composite numbers $N$ with two prime factors such that $N$ is a $K_{\alpha}$-number.
📌 3 Maddede Bu Makale
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Amaç & Hipotez:Let $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{Q}\setminus \{0\}$; a positive integer $N$ is said to be an \emph{$\alpha$-Korselt number} (\emph{$K_{\alpha}$-number}, for short) if $N\neq \alpha$ and $\alpha_{2}p-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $p$ of $N$.
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Temel Bulgular & Veri:In this paper we prove that for each squarefree composite number $N$ there exist finitely many rational numbers $\alpha$ such that $N$ is a $K_{\alpha}$-number and if $\alpha\leq1$ then $N$ has at least three prime factors.
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Sonuç & Katkı:Moreover, we prove that for each $\alpha\in \mathbb{Q}\setminus \{0\}$ there exist only finitely many squarefree composite numbers $N$ with two prime factors such that $N$ is a $K_{\alpha}$-number.
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:Prime number · Carmichael number · Korselt number
Abstract
Let $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{Q}\setminus \{0\}$; a positive integer $N$ is said to be an \emph{$\alpha$-Korselt number} (\emph{$K_{\alpha}$-number}, for short) if $N\neq \alpha$ and $\alpha_{2}p-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $p$ of $N$. In this paper we prove that for each squarefree composite number $N$ there exist finitely many rational numbers $\alpha$ such that $N$ is a $K_{\alpha}$-number and if $\alpha\leq1$ then $N$ has at least three prime factors. Moreover, we prove that for each $\alpha\in \mathbb{Q}\setminus \{0\}$ there exist only finitely many squarefree composite numbers $N$ with two prime factors such that $N$ is a $K_{\alpha}$-number.