Solve the following equations for a general $n$ such that:
(a) $ \phi(n) = \frac{n}{2} $
(b) $ \phi(n) = \frac{n}{3} $
We remind ourselves of Proposition (5.9) which states that for a natural number $n=p_1^{k_1} \times p_2^{k_2} \times \ldots \times p_r^{k_r}$, where $p_i$ are distinct primes, then
$$ \phi(n) = n \times \biggl(1- \frac{1}{p_1}\biggr) \times \biggl(1- \frac{1}{p_2}\biggr) \times \ldots \times \biggl(1- \frac{1}{p_r}\biggr) $$
(a) We require
$$ \frac{n}{2} = n \times \biggl(1- \frac{1}{p_1}\biggr) \times \biggl(1- \frac{1}{p_2}\biggr) \times \ldots \times \biggl(1- \frac{1}{p_r}\biggr) $$
This is only possible with $p_1=2$ and no other primes.
This means the general solution, for some natural number $k$, is
$$n = 2^k$$
(b) We require
$$ \frac{n}{3} = n \times \biggl(1- \frac{1}{p_1}\biggr) \times \biggl(1- \frac{1}{p_2}\biggr) \times \ldots \times \biggl(1- \frac{1}{p_r}\biggr) $$
This is only possible with $p_1=2, p_2=3$ and no other primes.
This means the general solution, for some natural numbers $j, k$, is
$$n = 2^j3^k$$