Dividers of a number n If $n=p_1^{2a_1}p_2^{2a_2}...p_n^{2a_n}$
Number of divisors of $n$:
$(2a_1+1)(2a_2+1)...(2a_n+1)$
I saw an other similar question, but it is complicated. There should be an more easy answer.

It's much more simple than that. For each divisor $d$ of $n$, $\frac nd$ is another divisor of $n$. So the divisors of $n$ come in pairs except if there is a divisor $d$ such that $d=\frac nd$. But this happens if and only if $n$ is a perfect square ($d=\frac nd\iff n=d^2$).
