Answer
$\phi(p^n)=p^n-p^{n-1}.\\
p\,\,is\,a\,prime\,and\,n\,is\,some\,positive\,integer\\
\phi(p^n)=p^n-p^{n-1}.\\
put\,\,p=2\,\,and\,let\,\,m=2n+1\\
\phi(2^m)=2^m-2^{m-1}=2^{m-1}(2-1)=2^{m-1}=2^{2n+1-1}=2^{2n}=(2^{n})^2\\
\because (2^{n})^2\,is\,a\,perfect\,sqaure.\\
\therefore \phi(2^m)is\,a\,perfect\,sqaure.\\
(there\,are\,infinitely\,many\,integers\,\,\,m)\\
\phi(n)is\,a\,perfect\,sqaure.(\,\,n=2^m)\\
$
Work Step by Step
$\phi(p^n)=p^n-p^{n-1}.\\
p\,\,is\,a\,prime\,and\,n\,is\,some\,positive\,integer\\
\phi(p^n)=p^n-p^{n-1}.\\
put\,\,p=2\,\,and\,let\,\,m=2n+1\\
\phi(2^m)=2^m-2^{m-1}=2^{m-1}(2-1)=2^{m-1}=2^{2n+1-1}=2^{2n}=(2^{n})^2\\
\because (2^{n})^2\,is\,a\,perfect\,sqaure.\\
\therefore \phi(2^m)is\,a\,perfect\,sqaure.\\
(there\,are\,infinitely\,many\,integers\,\,\,m)\\
\phi(n)is\,a\,perfect\,sqaure.(\,\,n=2^m)\\
$
