A note on solitary numbers
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arXiv
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| Format: | Preprint |
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2025
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| _version_ | 1866915496348090368 |
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| author | Mandal, Sagar |
| author_facet | Mandal, Sagar |
| contents | Does $14$ have a friend? Until now, this has been an open question. In this note, we prove that a potential friend $F$ of $14$ is an odd, non-square positive integer. $7$ appears in the prime factorization of $F$ with an even exponent while at most two prime divisors of $F$ can have odd exponents in the prime factorization of $F$. If $p\mid F$ such that $p$ is congruent to $7$ modulo $8$, then $p^{2a}\mid\mid F$, for some positive integer $a$. Further, no prime divisor of $F$ has an exponent congruent to $7$ modulo $8$ and no prime divisor can exceed $1.4\sqrt{F}$. The primes $3,5$ cannot appear simultaneously in the prime factorization of $F$. If $(3,F)>1$ or $(5,F)>1$, then $ω(F)\geq4$, otherwise $ω(F)\geq8$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2503_11694 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | A note on solitary numbers Mandal, Sagar General Mathematics 11A25 Does $14$ have a friend? Until now, this has been an open question. In this note, we prove that a potential friend $F$ of $14$ is an odd, non-square positive integer. $7$ appears in the prime factorization of $F$ with an even exponent while at most two prime divisors of $F$ can have odd exponents in the prime factorization of $F$. If $p\mid F$ such that $p$ is congruent to $7$ modulo $8$, then $p^{2a}\mid\mid F$, for some positive integer $a$. Further, no prime divisor of $F$ has an exponent congruent to $7$ modulo $8$ and no prime divisor can exceed $1.4\sqrt{F}$. The primes $3,5$ cannot appear simultaneously in the prime factorization of $F$. If $(3,F)>1$ or $(5,F)>1$, then $ω(F)\geq4$, otherwise $ω(F)\geq8$. |
| title | A note on solitary numbers |
| topic | General Mathematics 11A25 |
| url | https://arxiv.org/abs/2503.11694 |