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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.

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    $\begingroup$ It's easily checked that each of these is a divisor. You must show that they are the only divisors. Use the fact that $p|ab\implies p| a or p|b$ $\endgroup$ – saulspatz Apr 9 at 13:09
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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 $$ d=p_1^{a_1}p_2^{a_2}...p_r^{a_r} \mbox{ where for some } i \mbox{, } a_i>k_i$$

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