Say we have 2 prime numbers $a$,$b$ and $c=a*b$.

Is there any other pair of prime numbers $x,y$ (distinct from $a$ and $b$) so that $c=x*y$ ?

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    $\begingroup$ No. By unique factorization. $\endgroup$ – lulu Feb 10 '18 at 15:31
  • $\begingroup$ Welcome to Math SX! No, because the factorisation of a number is unique, up to the order of the factors. This is the Fundamental theorem of arithmetic. $\endgroup$ – Bernard Feb 10 '18 at 15:31
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    $\begingroup$ No. This is a consequence of unique prime factorization of the integers. It can be true however in some rings. $\endgroup$ – Quantaliinuxite Feb 10 '18 at 15:31
  • $\begingroup$ Apart from the obvious attempts $x=b$, $y=a$ (i.e., swapping - if $a\ne b$) or $x=-a$, $y=-b$ (if one happens to consider negatives of primes also primes) ... $\endgroup$ – Hagen von Eitzen Feb 10 '18 at 15:37

I've read and re-read your question and I see only one indication that you meant to restrict "prime numbers" to the prime numbers among the natural numbers (for this purpose it doesn't matter if you consider 0 a natural number or not).

That would be the fact that you used the tag , which has this official description:

Prime numbers are natural numbers greater than 1 not divisible by any smaller number other than 1. This tag is intended for questions about, related to, or involving prime numbers.

Let's assume that you did indeed mean such a restriction. Then the answer to your question is no. The natural numbers form a unique factorization domain (UFD) and we don't consider reordering to fundamentally change a factorization.

That means $ab$ is the same as $ba$, though in general we do prefer $ab$ if $a \leq b$. So for instance $2 \times 3$ and $3 \times 2$ are not distinct factorizations of 6. Well, you did say $x$ and $y$ have to be distinct from $a$ and $b$.

If you allow negative integers, it gets slightly more interesting. Without loss of generality, set $x = -a$ and $y = -b$. Then $ab = xy = c$. That's why it's better to say that ordering and multiplication by units (like $-1$) don't create distinct factorizations.

If you allow $c = 0$, things get a bit screwy. Then, if $a$ and $x$ are both 0, then $b$ and $y$ can be any integers at all as long $b \neq y$.

And if you look into rings of algebraic integers, many of which are not UFDs, things get much, much more interesting. But for the purpose of your question, are we sticking to the weaker definition of primality, which is now more commonly called "irreducible"? If so, the most famous example is $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$$ from $\mathbb Z[\sqrt{-5}]$. None of those numbers are primes by the stronger definition, but all the numbers to the right of the equals sign are irreducible in $\mathbb Z[\sqrt{-5}]$, they're not divisible by any numbers of smaller norm other than 1 and $-1$.

  • $\begingroup$ Nice answer, except that $\mathbb{N}$ is not a UFD, since $\mathbb{N}$ is not a ring (or domain), not being closed under subtraction. $\endgroup$ – vadim123 Feb 21 '18 at 22:58

Yes, there is, there are infinitely many in fact. Assuming we're talking about $\mathbb Z = \{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots \}$, we have, for example $$14 = 2 \times 7 = -2 \times -7$$ and $$-14 = -2 \times 7 = 2 \times -7.$$

These don't count as distinct factorizations because of multiplication by units. But the way you worded your question, unless you edit it, I think I have given a valid answer.

  • $\begingroup$ Prime numbers are usually defined to be positive. $\endgroup$ – robjohn Feb 11 '18 at 7:09
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    $\begingroup$ @robjohn That's more a lazy convenience for $\mathbb{Z}$ than anything else, it becomes almost a nuisance in other rings. Is $-2 + \sqrt{-7}$ a prime? What about $2 - \sqrt{7}$? $\endgroup$ – Bill Thomas Feb 12 '18 at 21:17
  • $\begingroup$ @BillThomas: indeed, if you want to properly include units and replace primes by prime equivalence classes in $\mathbb{Z}$, then $\{2,-2\}$, $\{3,-3\}$, $\{5,-5\}$, ... are prime classes, but unless you do, I would stick with the standard definition. $\endgroup$ – robjohn Feb 12 '18 at 21:27
  • $\begingroup$ @robjohn For what it's worth, Robert noted in his answer that the official tag description does say "natural numbers greater than 1". It's not clear if either Beach or Lisa have read that. $\endgroup$ – Bill Thomas Feb 12 '18 at 21:39

The answer is "sort of". We can factor $10$ into primes in four ways: $2\times 5$, $5\times 2$, $(-2)\times (-5)$, and $(-5)\times(-2)$. All of $2,5,-2,-5$ are prime.

However, in the integers, factorization into primes is unique up to (a) order, and (b) multiplication by $-1$. That is, if we rearrange $5\times 2$, we get $2\times 5$. If we multiply each of $-2, -5$ by $-1$, we turn $(-2)\times (-5)$ into $2\times 5$. This is called the Fundamental Theorem of Arithmetic.

If, instead of "integers", we consider different types of numbers, then different properties hold. The integers are what's called a UFD; in all UFD's a version of the fundamental theorem of arithmetic holds.

  • $\begingroup$ Prime numbers are usually defined to be positive. $\endgroup$ – robjohn Feb 11 '18 at 7:11
  • $\begingroup$ @robjohn The Wikipedia article you link is "semi-protected," so maybe pranksters can't edit it to contradict what you've asserted. $\endgroup$ – Mr. Brooks Feb 17 '18 at 22:13
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    $\begingroup$ @Mr.Brooks: I'd be surprised if anyone cared enough about my post to try changing the Wikipedia article just to contradict my answer. $\endgroup$ – robjohn Feb 18 '18 at 0:31
  • $\begingroup$ @robjohn People on the Internet can be that petty and lacking in perspective. $\endgroup$ – Mr. Brooks Feb 21 '18 at 22:02


$c = a*b = x*y$ would mean $a = \frac {x*y}{b}$. As $b$ is prime that would mean either $b|x$ or $b|y$. As $x,y$ are prime that would mean $b=x$ or $b = y$.

This is the Unique Factorization Theorem also called the Fundamental Theorem of Algebra.

Notice, however I took for granted, without proof, that if a prime $b$ divides $x*y$ then $x$ must divide either $x$ or $y$. This assumption is Euclid's Lemma and is key to proving Unique Factorization Theorem.


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