Show there are infinitely many positive integers $k$ such that $\phi(n)=k$ has exactly two solutions where $n$ is a positive integer 
Show that there are infinitely many positive integers $k$ such that the equation $\phi(n)=k$ has exactly two solutions, where $n$ is a positive integer.

Not entirely sure where to start or finish with this problem. I know that $\phi$ notation states that $\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_S})$
but other than that I am lost.

Let $k = 2 \times 3^{6m + 1}$. We claim that if $m > 0$, then $\phi(n) = k$ has exactly $2$ solutions. (For $m = 0$, it has four solutions: $7$, $9$, $14$, and $18$.)
Let $n = p_1^{a_1} p_2^{a_2} \cdots p_i^{a_i}$, where $p_j$ is a prime number, and $a_j$ is a positive natural number. Then $n$ is a solution if and only if
$$
  p_1^{a_1 - 1} p_2^{a_2 - 1} \cdots p_i^{a_i - 1} (p_1 - 1)(p_2 - 1) \cdots (p_i - 1) = k.
$$
We note that if $n$ has at least two odd prime factors $p$ and $q$, then $(p - 1)(q - 1)$ is a factor of $\phi(n)$. Hence $4$ is a factor of $\phi(n)$, and sol $n$ is not a solution.
Thus any solution $n$ is of the form $n = 2^\alpha \cdot p^\beta$ where $p$ is some odd prime number, and $\alpha$ and $\beta$ are natural numbers, possibly equal to $0$. If $\alpha > 2$, then $\phi(8) = 4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus $\alpha \in \{0, 1, 2\}$.
Suppose that $\alpha = 2$. Then if $\beta > 0$, we have that $2(p - 1)$ is a factor of $\phi(n)$, and so $4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus if $\alpha = 2$, then $\beta = 0$, and so $n = 4$. But $\phi(4) = 2$ is not divisible by $3$, and so it is not equal to $2 \times 3^{6m + 1} = k$.
We thus must have that either $n = p^\beta$, or $n = 2 p^\beta$. In each case, we have that
$$
  \phi(n) = p^\beta - p^{\beta - 1}.
$$
We thus must solve the equation
$$
  p^{\beta - 1} (p - 1) = 2 \times 3^{6m + 1}.
$$
If $\beta = 1$, then this is equivalent to $p - 1 = 2 \times 3^{6m + 1}$. Not we note that $2 \times 3^{6m + 1} + 1 \equiv 2 \times 3 + 1 \equiv 0 \pmod 7$, and so this implies that $p = 7$. This corresponds to $m = 0$. We're considering the case where $m > 0$, so we can thus assume from now on that $\beta > 1$.
Since $\beta > 1$, we note that $p \mid 2 \times 3^{6m + 1}$, and so we have that $p = 3$. We thus must solve the equation
$$
  2 \times 3^{\beta - 1} = 2 \times 3^{6m + 1}.
$$
This clearly has the unique solution $\beta = 6m + 2$, and so the two solutions to the equation $\phi(n) = k$ are given by $n = 3^{6m + 2}$, and $n = 2 \times 3^{6m + 2}$.
