I understand the reason 1 is not considered to be a prime number, but what is the reasoning for -1 not being considered a prime number?

It's only factors are 1 and itself, -1, wouldn't that make it a prime number?

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    $\begingroup$ Primes are defined as positive integers. There's not much more reasoning than that. $\endgroup$ – SeraPhim Jul 10 '20 at 21:50
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    $\begingroup$ From a ring-theory standpoint, $-1$ is not a prime in the ring of integers because it is a unit, i.e. it has a multiplicative inverse (namely itself). $\endgroup$ – Robert Israel Jul 10 '20 at 21:53
  • $\begingroup$ Also relevant: math.stackexchange.com/questions/1645111/… $\endgroup$ – an4s Jul 10 '20 at 21:55
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    $\begingroup$ John Horton Conway, who died recently, started to regard $-1$ as a prime because of concerns in the number theory of integer quadratic forms. Granted, this was probably written down only in his book The Sensual Quadratic Form. $\endgroup$ – Will Jagy Jul 10 '20 at 22:15
  • $\begingroup$ @WillJagy, a pertinent excerpt is at math.stackexchange.com/questions/1175367/is-the-number-1-prime/… (which I found through the link in an4s's second comment). $\endgroup$ – Barry Cipra Jul 10 '20 at 22:16

John Horton Conway came up with an alternative to Jordan decomposition for describing integer quadratic forms. It uses his version of $p$-adic symbols, including one for $p=-1.$ Notice that this is primarily a replacement for usage of $\infty$ as a prime, which we see in the excellent book of Cassels, Rational Quadratic Forms.

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