Negative factors of a number Can a factor of a number be negative?
Is $-5$ a factor of $25$ or $-25$?
A number is said to be prime if it has two factors : $1$ and the number itself. So if $-5$ can be a factor of $5$, how to define the prime numbers?
Thanks.
 A: I think for the most part in primary and secondary school, we stick to factoring positive integers to sidestep this issue. We deal with it in higher mathematics though!
Technically, when factoring integers, factors that differ only by $\pm 1$ are considered to be the same factor.
So really, $25=5\cdot5=(-5)\cdot(-5)$ are both factorizations of 25, but we consider them to be "the same" because the factors only differ from each other by $-1$. By convention, we usually just talk about the positive versions as being "the factors."
In ring theory, the significance of prime numbers in $\Bbb Z$ is that the ideal $(a)\lhd\Bbb Z$ is a prime ideal iff $a$ is a prime number. Since $(5)=(-5)$ as ideals, then it seems fair that both 5 and -5 are considered as "prime."
For principal ideal domains more complicated than $\Bbb Z$, it gets even more interesting. If $p$ is a prime in a domain $R$ and $u$ is a unit of the ring, then $pu$ and $p$ are considered them same prime, since $(pu)=(p)$ is a prime ideal. This applies to the integers because in $\Bbb Z$, the only units are $\pm 1$.
A: Indeed, when we enlarge from positive integers to all integers then we have to deal with the fact that this introduces a nontrivial unit (invertible) element $\,-1,$ which may spoil the uniqueness of prime factorizations (up to order), since, e.g. $\,pq = (-p)(-q)\,$ by $\,(-1)(-1) = 1.$  Similarly for polynomials with rational coefficients. Here every nonzero rational number $\,r\,$ is a unit  (invertible) $\ r r^{-1} = 1,\,$ so $\,p(x) q(x) =  (r p(x)) (r^{-1} q(x)).\,$  There are various ways to remedy this. 
First, from  the set of all unit multiples of a prime, we could choose some "natural" representative, and call only that element prime. For integers, a natural choice is the positive representative, and for polynomials one that is monic  (leading coefficient $= 1).$ Such representatives are sometimes called unit-normalized representatives.
Alternatively, we may permit all unit multiples to be prime, then alter our statement of uniqueness of prime factorizations to say: prime factorizations are unique up to order and unit multiples. 
If you study abstract algebra you will learn that there is a natural algebraic way of "ignoring" these pesky nontrivial units when studying factorization. Namely, since the units form a group $\,U,\,$ we can mod out by $\,U.\,$ This is the algebraic way of working multiplicatively "up to unit multiples". 
A: You raise an interesting question, which has repercussions in various forms of more advanced algebra. When we are dealing with the positive integers there is no difficulty with definitions - a prime is a positive integer, but in wider contexts there may be no natural order to say whether a number is positive or negative. For example $i=\sqrt {-1}$ cannot be distinguished algebraically from $-i$.
The one point of interest is whether we should count $1$ as a prime - and the general convention is that it is counted not as a prime, but a unit. $-1$ is also a unit - we can add $(-1)^2$ to any factorisation without changing its value.
Numbers which are obtained from each other by multiplication by a unit are often known as associates, so $-5, 5$ are associates in the integers, and it can be proper to call them both (associated) primes.
There are certain things we can only do "up to multiplication by units" - and factorisation and the identification of prime numbers are two of these.
However, back to the beginning - with the integers, associated primes come in pairs, and it is natural and convenient to work with the positive member of each pair.
A: The condition you have given does not suffice to make a number prime. It is also generally required that it be greater than $1$, and that the divisors considered be positive.
