Number of Divisors of a Square Fre Number

  is square free.  
  is not divisible by any square number or the square of any prime number.
  must be of the form  
  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  
\[\tau (n)=2^m\]
  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.

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