# How to prove $d>0$ is a divisor of $n$ iff $d=p_1^{b_1}p_2^{b_2}…p_r^{b_r}$ with $0< b_i<a_i$ for each $i$?

How to prove $d>0$ is a divisor of $n$ iff $d=p_1^{b_1}p_2^{b_2}...p_r^{b_r}$ with $0< b_i<a_i$ for each $i$? here $n = p_1^{a_1}p_2^{a_2}...p_r^{a_r}$ with the $p_i$ distinct primes and the $a_i$ positive integers.

I am unsure of how to start this problem any solutions or hints are welcome

• You want $0\le b_i\le a_i$. This is really just unique factorisation into primes. – Lord Shark the Unknown Oct 19 '17 at 15:54
• Factor each $p_i^{a_i}= p_i^{b_i}p_i^{a_i- b_i}$.. Since "$0< b_i< a_i$" $a_i- b_i$ is positive. – user247327 Oct 19 '17 at 15:55

Assume $d = p_1^{b_1}p_2^{b_2}...p_r^{b_r}p_{r+1}^{b_{r+1}}$ (i.e. $d$ has a prime factor that is not a factor of $n$). Divide $n$ by $d$.
Then, assume $d = p_1^{b_1+k}p_2^{b_2}...p_r^{b_r}$ (i.e., $d$ has one of the prime factors raised to a power higher than what it was raised in $n$). Again, divide $n$ by $d$.