# Positive divisors of prime factorization of n

Let $$n=p_1^{k_1}p_2^{k_2}...p_r^{k_r}$$ be the prime factorisation of $$n>1$$. Show that the positive divisors of n are the integers $$d=p_1^{a_1}p_2^{a_2}...p_r^{a_r}$$ for $$i=1,2,...,r$$, with $$0\le a_i\le k_i$$.

Attempt: We know the positive divisors of n that are powers of $$p_1$$ are $$1, p_1, p_1^2, p_1^3, ..., p_1^{k_1}$$, and for $$p_2$$ are $$1, p_2, p_2^2, p_2^3, ..., p_2^{k_2}$$ and so on up to $$p_r$$. Could $$d(n)$$, the number of divisors function be used somehow? Not really sure how to approach this.

• It's easily checked that each of these is a divisor. You must show that they are the only divisors. Use the fact that $p|ab\implies p| a or p|b$ – saulspatz Apr 9 at 13:09

You could also take this approach. Also remember the definition of a prime ($$p|ab \implies p|a \mbox{ or } p|b$$)
• First, show that any integer of the form $$d=p_1^{a_1}p_2^{a_2}...p_r^{a_r}$$ for $$i=1,2,...,r$$, with $$0\le a_i\le k_i$$ does divide $$n$$ (this is clear and direct..)
• Second you can show that any integer not of that form does not divide $$n$$. To do this, there are 2 cases: $$d \mbox{ has a prime factor } q\neq p_i \mbox{ for any }i$$ or $$d=p_1^{a_1}p_2^{a_2}...p_r^{a_r} \mbox{ where for some } i \mbox{, } a_i>k_i$$