Number of Divisors of a Square Fre Number
\[n\]
is square free. \[n\]
is not divisible by any square number or the square of any prime number.\[n\]
must be of the form \[n=p_1p_2...p_k\]
then using Euler's Totient Function the number of divisors is \[\tau (n)=\underbrace{(1+1)...(1+1)}_{k \; times}=2^k\]
.The converse is not true. That is, if
\[\tau (n)=2^m\]
then \[n\]
need not be square free. For example, \[\tau (135)=\tau (3^3 \times 5)(1+3)(1+1)=2^3\]
but 135 is divisible by 9. 
