More precisely, we prove that if $(\mathbb{Q\setminus}\mathbb{Z})$-$\mathcal{KS}(pq)=\emptyset$, then $\mathbb{Z}$-$\mathcal{KS}(pq)=\{ q+p-1\}$.
📌 3 Maddede Bu Makale
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Amaç & Hipotez:For a positive integer $N$ and $\mathbb{A}$, a subset of $\mathbb{Q}$, let $\mathbb{A}$-$\mathcal{KS}(N)$ denote the set of $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{A\setminus} \{0,N\}$, where $\alpha_{2}r-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $r$ of $N$.
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Temel Bulgular & Veri:In this paper, we prove that each $pq$-Korselt base in $\mathbb{Z\setminus}\{ q+p-1\}$ generates at least one other in $\mathbb{Q}$-$\mathcal{KS}(pq)$.
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Sonuç & Katkı:More precisely, we prove that if $(\mathbb{Q\setminus}\mathbb{Z})$-$\mathcal{KS}(pq)=\emptyset$, then $\mathbb{Z}$-$\mathcal{KS}(pq)=\{ q+p-1\}$.
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:prime number · Carmichael number · squarefree composite number
Abstract
For a positive integer $N$ and $\mathbb{A}$, a subset of $\mathbb{Q}$, let $\mathbb{A}$-$\mathcal{KS}(N)$ denote the set of $\alpha=\dfrac{\alpha_{1}}{\alpha_{2}}\in \mathbb{A\setminus} \{0,N\}$, where $\alpha_{2}r-\alpha_{1}$ divides $\alpha_{2}N-\alpha_{1}$ for every prime divisor $r$ of $N$. The set $\mathbb{A}$-$\mathcal{KS}(N)$ is called the set of $N$-Korselt bases in $\mathbb{A}$. Let $p, q$ be two distinct prime numbers. In this paper, we prove that each $pq$-Korselt base in $\mathbb{Z\setminus}\{ q+p-1\}$ generates at least one other in $\mathbb{Q}$-$\mathcal{KS}(pq)$. More precisely, we prove that if $(\mathbb{Q\setminus}\mathbb{Z})$-$\mathcal{KS}(pq)=\emptyset$, then $\mathbb{Z}$-$\mathcal{KS}(pq)=\{ q+p-1\}$.
Ghanmi, N. (2022). Connections on the rational Korselt set of $pq$. Hacettepe Journal of Mathematics and Statistics, Cilt: 50 - Sayı: 1, 135–143. https://ajindex.com/articles/529441