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Let  
\[n=p_1^{k_1}p_2^{k_2}...p_r^{k_r}\]
  then  
\[n^2=p_1^{2k_1}p_2^{2k_2}...p_r^{2k_r}\]
, where  
\[\phi (n)\]
  is the number of integers less than or equal to  
\[n\]
  which are relatively prime to  
\[n\]
.
\[\begin{equation} \begin{aligned} \phi (n^2) &= \phi (p_1^{2k_1}p_2^{2k_2}...p_r^{2k_r}) \\ &= \phi (p_1^{2k_1}) \phi (p_2^{2k_2}) ... \phi (p_r^{2k_r})  m(1)) \\ &= p_1^{2k_1-1} (p_1-1)p_2^{2k_2-1} (p_2-1)... p_r^{2k_r-1} (p_r-1) \\ &= (p_1^{k_1}p_2^{k_2}...p_r^{k_r}) (p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)) \\ &= n \phi (n) \end{aligned} \end{equation}\]
.
(1) is used since  
\[\phi (ab)= \phi (a) \phi (b)\]
  whenever  
\[a, \; b\]
  are relatively prime.
Also since  
\[\phi (p^k)=p^{k-1}(p-1)\]
  then
\[\begin{equation} \begin{aligned} \phi ( \phi (p^k)) &= \phi (p^{k-1}(p-1)) \\ &= \phi (p^{k-1}) \phi (p-1) \\ &= p^{k-2}(p-1) \phi (p-1) \\ &= p^{k-2} \phi ((p-1)^2) \end{aligned} \end{equation}\]

on using the first result.