I saw this somewhere and remembered it when presented with a question concerning the number of positive integers that are divisors for some integer.

Consider :

For some prime factorization: $$a^{i}*b^{j}*c^{k}*...z^{n}=d$$ $d \in\mathbb{Z}$ and $i,j,k...n \in\mathbb{N_0}$

then $(i+1)(j+1)(k+1)(n+1)$ yields the number of all positive divisors of $d$.

Is there a proof by induction for this? Is this an already existing theorem or lemma that I might have missed?


Yeah, you can do induction on the number of distinct prime divisors.

If $n=p^k$ then the divisors of $n$ are $1, p, \dots, p^k$, so there are $k+1$ of them.

$If n=p_1^{k_1} \dots p_{r+1}^{k_{r+1}}$ then by induction hypothesis, $n$ has $(k_1+1) \dots (k_r + 1)$ divisors $d$ coprime with $p_{r+1}$, and for each of these $d$ you get exactly $k_{r+1}+1$ distinct divisors of $n$, namely $d, p_{r+1}d, \dots, p_{r+1}^{k_{r+1}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.