Show that $\phi (2^{2k+1}) = l^2$ where $l$ is a natural number.
We have
$$ \phi(2^{2k+1}) = 2^{2k +1 -1 }(2-1) = (2^{k})^2 $$
And so, $\phi (2^{2k+1}) = l^2$ where $l=2^k$.