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Let $n$ be a positive integer.

Can $n^2-1$ be a Carmichael number ?

Some thoughts :

  • n must be even, since $n^2-1$ must be odd.
  • Every prime factor $p\mid n^2-1$ must satisfy $p-1\mid n^2-2$. This rules out prime factors of the forms $3k+1,4k+1,5k+1$
  • Neither $n-1$ nor $n+1$ can be prime. In the first case, we have $n-2\mid n^2-2$ , implying $n-2\mid 2$ , hence $n-2\le 2$. In the second case , we have $n\mid n^2-2$ , hence $n\mid 2$, hence $n\le 2$. But the smallest Carmichael number is $561$, hence $n$ must at least be $24$.
  • Brute force reveals that for $n\le 10^9$, there is no Carmichael number of this form.
Peter
  • 86,576
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    For the case $n=2^k$ there are some posts, e.g., here, but I don't know how helpful they are. You checked until $2^{3500}-1$ several years ago. November $5$ of $2015$, to be precise. – Dietrich Burde Mar 29 '21 at 08:41
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    I extended my search limit to $10^{10}$ – Peter Mar 29 '21 at 10:29
  • A thought toward a partial solution: Having no prime factors of the form $3k+1$ means all prime factors are either $3$ or of the form $6k-1$. Setting aside the possibility of $3$ for the moment, if $n-1$ and $n+1$ have only factors of the form $6k-1$, then they themselves are of the form $6m\pm 1$. This would require that $n-1$ has an odd number of prime factors, and $n+1$ has an even number of prime factors. – Keith Backman Mar 29 '21 at 15:01
  • Antoher thought: $n^2-1$ has to be a pseudo-prime to base $2$, since all Carmichael numbers fulfill this condition. – Martin Hopf Mar 29 '21 at 15:21

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