A note on necessary conditions for a friend of 10
Tapas Chatterjee, Sagar Mandal, Sourav Mandal
TL;DR
This work studies the problem of identifying friends of $10$, i.e., integers $N>10$ with $I(N)=I(10)=\dfrac{9}{5}$, by refining necessary conditions on prime factorization and divisibility of divisor sums. Building on Ward’s results, the authors prove that any friend of $10$ must have at least seven distinct odd prime factors and must contain primes with specific congruence properties, notably $p\equiv1\pmod{10}$ and $q\equiv1\pmod{6}$, and they introduce the notation $f_{P}^{Q}$ to capture orders governing $p|\sigma(q^{2a})$. A sequence of results (Theorems 1.3, 1.7, 1.2, 1.9, 1.10 and Corollaries 1.4, 1.5, 1.6, 1.8, 1.11) develops deep structural constraints on $N$, yielding lower bounds on the total number of prime factors and, in several cases, upper bounds on potential growth. The analysis relies on the abundancy index’s weak multiplicativity, valuations of divisor sums, and intricate congruence arguments, and it shows that if a friend of $10$ exists, it must be extraordinarily large; however, the exact existence remains open. The paper also provides a comprehensive catalog of the $f_p^{q}$ values used in the proofs and discusses implications for Fermat primes and the $N=5^{2a}m^2$ decomposition.
Abstract
Solitary numbers are shrouded with mystery. A folklore conjecture assert that 10 is a solitary number i.e. it has no friends. In this article, we establish that if $N$ is a friend of $10$ then it must be odd square with at least seven distinct prime factors, with $5$ being the least one. Moreover there exists a prime factor $p$ of $N$ such that $2a+1\equiv 0 \pmod f$ and $5^{f}\equiv 1 \pmod p$ where $f$ is the smallest odd positive integer greater than $1$ and less than or equal to $\min\{ 2a+1,p-1\}$, provided $5^{2a}\mid \mid N$. Further, there exist prime factors $p$ and $q$ (not necessarily distinct) of $N$ such that $p\equiv1 \pmod {10}$ and $q\equiv 1\pmod 6$. Besides, we prove that if a Fermat prime $F_k$ divides $N$ then $N$ must have a prime factor congruent to $1$ modulo $2F_k$. Also, if we consider the form of $N$ as $N=5^{2a}m^2$ then $m$ is non square-free. Furthermore, we show that $Ω(N)\geq 2ω(N)+6a-4$ and if $Ω(m)\leq K$ then $N< 5\cdot 6^{(2^{K-2a+1}-1)^2}$ where $Ω(n)$ and $ω(n)$ denote the total number of prime factors and the number of distinct prime factors of the integer $n$ respectively.