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This question already has an answer here:

We all know that $$15=3 \times 5$$

And $$15 =(-3) \times(-5)$$

Since $3 \neq -3$ and $5 \neq -5$ , we have two different prime factorizations !

Is this wrong ?

If this is wrong , then there are no negative primes !

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marked as duplicate by Daniel Fischer Jan 25 '16 at 11:54

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    $\begingroup$ Primes are defined as integers greater than or equal to two that are only divisible by positive integers $1$ and themself. $\endgroup$ – Gregory Grant Jan 25 '16 at 10:01
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    $\begingroup$ When properly stated, and there is unique factorization, the uniqueness is only up to multiplication by a unit in the ring. In the ring $\mathbb{Z}$, the units are $\pm 1$. The factorization of a natural number greater than one as a product of prime positive integers is absolutely unique ( but $\mathbb{N}$ is no longer a ring). In the Gaussian integers $\mathbb{Z}[i]$, which has unique factorization, the units are $\pm 1, \pm i$. $\endgroup$ – Geoff Robinson Jan 25 '16 at 10:01
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    $\begingroup$ It depends on the definition of primes that you use. 1) $p$ is prime if $p>0$ and it has exactly the two divisors $1,p$. 2) $p$ is prime if $p$ and it has exactly the four divisors $1,p,-1,-p$. For example, for Def2, Fundamental Therorem claims that every integer different for 0, 1 and -1 has a unique prime factorization unless order and sign. $\endgroup$ – sinbadh Jan 25 '16 at 10:06
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    $\begingroup$ This is the kind of question I personally find extremely uninteresting. It's just a minor issue with definitions that gets in the way of more interesting/important ideas. $\endgroup$ – Zubin Mukerjee Jan 25 '16 at 10:07
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See here, here (math.SE) and here (math.SE)

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