# Proof for formula for number of divisors

For any natural number $$n>1$$, with factorization $$n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m}$$ we can find the number of divisors by using the formula, $$(\alpha_1+1)(\alpha_2+1)....(\alpha_m+1)$$ How do I find this formula. Is there a proof for it?

• Do you know multiplicative functions? – s1mple Apr 12 at 12:55
• Try it for some small values of $\alpha$. ... say $n=6$. – Donald Splutterwit Apr 12 at 12:57
• @s1mple how do i use them as a proof? – Kirito Apr 12 at 13:00
• @Kirito please see the explanation. – s1mple Apr 12 at 13:02

To prove this first consider the number of the form $$n=p^{\alpha}$$. The divisors are $$1,p,p^2,\cdots, p^{\alpha}$$, i.e. $$d(p^{\alpha})=\alpha+1$$. Now, consider $$n=p^{\alpha}q^{\beta}$$, where $$p,q$$ are co-prime. The divisors would be : $$1,p,p^2,\cdots p^{\alpha}$$
$$q,pq,p^2q,\cdots,p^{\alpha}q$$ Similarly, upto: $$q^{\beta},pq^{\beta},\cdots,p^{\alpha}q^{\beta}$$ In this case, $$d(p^{\alpha}q^{\beta})=(\alpha+1)(\beta+1)$$. Hence $$d(n)$$ is multiplicative function.
For any natural number $$n>1$$, with prime factorization , where $$q_1,q_2\cdots q_m$$ are distinct primes and $$m\ge1$$ $$n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m}$$ we can find the number of divisors by using the formula, $$(\alpha_1+1)(\alpha_2+1)....(\alpha_m+1)$$