Euler's Totient Function is Odd for a Square Number

Euler's Totient Function,  
\[\tau (n)\]
  which returns the number of divisors of an integer  
\[n\]
  is odd if and only if  
\[n\]
  is a square number.
Suppose  
\[n\]
  is square number then  
\[n=p_1^{2k_1}p_2^{2k_2}...p_r^{2k_r}\]
.
\[\tau\]
  is multiplicative so that if  
\[r, \; s\]
  are relatively prime, then  
\[\tau (rs)=\tau (r) \tau (s)\]
  hence
\[\begin{equation} \begin{aligned} \tau (n) &= \tau (p_1^{2k_1}) \tau (p_2^{2k_2})... \tau (p_r^{2k_r}) \\ &= (1+2k_1)(1+2k_2)... (1+2k_r) \end{aligned} \end{equation}\]

The last expression is odd since each factor is odd. If  
\[\tau (n)\]
  is even then at least one factor  
\[(1+2k_s)\]
  is even bu this is impossible since  
\[k_s\]
  is an integer.

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