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For starters, I am not a mathematician, but I was fascinated by prime numbers for many years. So today, I came up with the definition which I am extremely curious about, because I believe it poses a mathematical problem.

Prime number is a positive integer that has exactly two positive integer divisors.

Here, I assume that if p is a positive integer, then 1 and p are among its divisors.

So the question, is there a positive integer number which is not prime and which has exactly two positive integer divisors?

In case it is already known definition, I would be thankful if you pointed me to it.

Arislan Makhmudov
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  • Why do you think the definition "poses a mathematical problem"? Try it with $n=1,2,3,4,\ldots$. A positive prime $p$ can only be factored as $1\cdot p=p\cdot 1$ in positive integers, since it is irreducible in $\Bbb Z$. In this ring, $p$ is prime if and only if $p$ is irreducible. – Dietrich Burde Oct 15 '21 at 12:03
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    I don't understand. The definition you have for a prime is precisely that it has exactly two positive divisors. Therefore, any natural number that has exactly two positive divisors is prime. – lulu Oct 15 '21 at 12:05
  • Nope. Exactly two divisors is the very definition of a prime. What you are asking is like "is there a positive number less than zero" ? –  Oct 15 '21 at 12:05
  • @lulu He says that this definition is false, i.e., is a "problem". – Dietrich Burde Oct 15 '21 at 12:05
  • Not possible. If $n > 1$ has prime factorization of $$\prod_{i=1}^r (p_i)^{(a_i)}$$ then the number of divisors of $n$ (including $n$ and $1$) is $$\prod_{i=1}^r (a_i + 1).$$ The only way that this evaluates to $(2)$ is if the number of distinct prime factors is $(r = 1)$ and if $a_1 = 1.$ – user2661923 Oct 15 '21 at 12:06
  • @DietrichBurde Ah, thanks. Though of course there is nothing wrong with the definition. – lulu Oct 15 '21 at 12:07
  • @DietrichBurde I think it's a mathematical problem because the definition doesn't say anything about 1 and p, in theory it could different numbers. – Arislan Makhmudov Oct 15 '21 at 12:08
  • @user2661923: if there are non-primes having two divisors, the prime factorization property falls apart. I am afraid that your sophisticated argument does not work. –  Oct 15 '21 at 12:09
  • Any natural number $n>1$ is divisible by itself and $1$. That's two divisors right there. If $n=ab$ is composite (so $1<a,b<n$) then $a,b$ would be more divisors. – lulu Oct 15 '21 at 12:10
  • You should have said that in the question ! As any number has $1$ and itself for divisors, it is implicit that a prime has no other divisor. –  Oct 15 '21 at 12:10
  • @YvesDaoust then we have another definition for prime numbers, of course if it doesn't already exist. – Arislan Makhmudov Oct 15 '21 at 12:10
  • @YvesDaoust Interesting. I assumed that one could take as a premise the prime factorization theorem, and then use the formula for number of divisors that is taught in my Number Theory book. So, if I understand correctly, I am guilty of circular reasoning. This is interesting, because I was simply quoting the Number Theory book. If you can't use that as a basis for analysis, what can you use? – user2661923 Oct 15 '21 at 12:12
  • @YvesDaoust but I have said it:

    Here, I assume that if p is a positive integer, then 1 and p are among its divisors.

     – Arislan Makhmudov Oct 15 '21 at 12:13
  • @YvesDaoust May be it's a stupid argument then that I am using in the question. But I am not a mathematician, so I beg your pardon. – Arislan Makhmudov Oct 15 '21 at 12:14
  • This isn't a terrible question for a non-mathematician. – Randall Oct 15 '21 at 12:18
  • @ArislanMakhmudov: see my answer. –  Oct 15 '21 at 12:19
  • @user2661923: well, you could use as a definition "a prime number is a number that belongs to the subset of naturals that ensures the uniqueness of the decomposition of a natural as a product of naturals" but that would be paranoid. Then you would indeed have to prove that a a prime has exactly two divisors. –  Oct 15 '21 at 12:23
  • @Randall, thank you! To be honest, the point was to propose a "new" definition of prime number, if it's really new and can not be overturned. From the discussion here I can see that it can't be. YvesDaoust pointed out that 1 and p are already divisors of p, so this makes two. I somehow missed this point, and it was stupid, otherwise I would have posed the question differently. So can it be considered an alternative definition? – Arislan Makhmudov Oct 15 '21 at 12:25

2 Answers2

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The number $1$ has exactly one positive integer divisor, namely $1$.
If $a$ is prime, then it has exactly two positive integer divisors, namely $1$ and $a$.
If $a$ is not $1$ and not prime, then it has at least three positive integer divisors: $1$ and $a$ and at least one other.

The definition in the OP:

Prime number is a positive integer that has exactly two positive integer divisors.

tells us the meaning of the word prime for this purpose.

Why is the definition stated like this? Long ago, the definition may have been: "A number $a$ not divisible by any positive integer except $1$ and $a$." This definition would include $1$ as a prime. When mathematicians found it more useful not to include $1$ among the primes, they came up with the defintion in the OP.

GEdgar
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A natural number always has the two divisors one and itself. So it saying "exactly two divisors" and "only the divisors one and itself" are equivalent statements. The first is more compact.

  • You have answered my question as well, but the first one answers my question too and was posted earlier, so I decided to choose it as correct. Thank you so much! – Arislan Makhmudov Oct 15 '21 at 12:40
  • @ArislanMakhmudov: in fact, gedgar's answer is better as it correctly deals with $1$. –  Oct 15 '21 at 12:43