# Divisibility and factors [duplicate]

1) Can factors be negative? Please prove your opinion. 2)If prime factorization is given to you, how will you find out how many composite factors are there? Not the factors, just how many.

For 2), my teacher gave me an easy method:

let x = $a^{a_1} . b^{b_1}. c^{c_1}..$

Then your answer is : $(a_1 +1)(b_1 + 1)(c_1 +1)...$

But again, he didnt prove it, and I didnt have any time left in the class to ask him, so please help me prove it...

• Your second sentence is weird. One cannot prove an opinion. – Tobias Kildetoft May 30 '13 at 16:39
• @Saurabh, you keep using that word "opinion". I don't think it means what you think it means. – J. M. isn't a mathematician May 30 '13 at 16:43
• Focus. How about answering my question. Don't use the formula your teacher gave you. Try to discover it for yourself by doing examples. This is one of the essential skills needed to develop as a mathematical thinker. – Ted Shifrin May 30 '13 at 16:54
• Remark that this depends crucially on the uniqueness of prime factorizations, i.e. the fundamental theorem of arithmetic. So be sure to mention that somewhere in your proof. – Key Ideas May 30 '13 at 17:18
• See math.stackexchange.com/questions/404783/… for comments on your uestion about negative factors – Mark Bennet May 30 '13 at 17:23

1) Prime factors are certainly positive only, since prime numbers are positive by definition. But a divisor of a number can certainly be negative. For example, all numbers in the set $\{-10,-5,-2,-1,1,2,5,10\}$ divide $10$.
2) Say $x = p_1^{e_1} * p_2^{e_2} *p_3^{e_3} * \dots * p_n^{e_n}$, where $p_i$ is the $i$th prime factor of $x$ and $e_i$ the exponent of that factor. If $a$ divides $n$, then all prime factors of $a$ have to be in the prime factorisation of $x$, and the exponent of such a factor in the factorisation in $a$ can be no more than the exponent of that factor in the factorisation of $x$. Therefore, the exponent of $p_1$ in the prime factorisation of $a$ is in the set $\{0,1,2,\dots , e_1\}$. More specifically, it is one of $e_1 + 1$ different values. Similarly, the exponent of $p_2$ in the prime factorisation of $a$ can take up $e_2 + 1$ values. Multiplying all these numbers together shows that the number of (positive) divisors of $a$ is indeed $(e_1+1)*(e_2+1)*(e_3+1)*\dots *(e_n+1)$.
For example, if $x = 12 = 2^2 * 3^1$, a divisor $a$ of $x$ can have a factor $2$ in its prime factorisation, two of them, or zero of them. Three cases for the factor $2$. Also, it can either have or have not a factor $3$ in its prime factorisation. Two cases for the factor $3$. Therefore, the possible total number of values $a$ can have is $3 * 2 = 6$ and indeed this is true: the positive factors of $12$ are $1,2,3,4,6,12$.
As I said above, a divisor can also be negative. If you want to count not only the number of positive divisors but the number of positive and negative divisors combined, you just have to multiply the outcome by $2$, because the number of positive and negative divisors are the same.