find the solutions of $\phi(n) = 14$. I tried the following.
Let $n = \Pi p_i^{\alpha_i}$, be a prime factorization of $n$, Then we can see that:
$\phi(n) = \Pi(p_i^{\alpha_i} - p_i^{\alpha_i-1}) = 14 = 2\cdot7$.
The main problem here is that i don't know how to find $p_i$ such that the last equation does hold or not. Is there a trick in finding these solutions?
Kees
 A: $$\phi(n)=\prod_{i=1}^m p_i^{\alpha_i-1}(p_i-1)=2\cdot 7\implies \exists p_i=7\ \mathrm{but}\ p_i-1=1\ \mathrm{or}\ 2$$ which leads to a contradiction, hence no such $n$ exists.
In fact, a little reflection reveals that $\phi(n)$ maybe square free only if $n$ is some power of a prime.
A: $2 \cdot7$ is the only non-trivial way to write $14$ as a product of two naturals.
Hence, you have to consider exactly the following cases ($p, q$ prime):
1.) $n = p^k$, i. e. $p^k - p^{k-1} = 14$
2.) $n = p^k q^m$, i. e. $p^k - p^{k-1} = 2 \wedge q^m - q^{m-1} = 7$
Case 1:
If $k = 1$, then $p^k - p^{k-1} = 14$ has the solution $p=15$, which is not prime. If $k \ge 2$, $14$ must divide the prime or $p - 1$. $14|p$ is impossible. If $14|(p-1)$, then $p \ge 15$. But we have $p^k - p^{k-1} \ge p^2 - p$ (since $p^k - p^{k-1}$ grows if $k$ grows), and
$$
p^2 - p = p(p-1) \ge 15 \cdot 14 > 2.
$$
Case 2:
Let's first determine $q$ and $m$. $m=1$ is impossible, since $q - 1 = 7$ has solution $q=8$, which is not prime. If $m \ge 2$, then $7$ must divide either $q$ or $q-1$. $q = 7$ is impossible, since $7^k - 7^{k-1} \ge 49 - 7 > 7$.
We conclude that 14 is not in the image of $\phi$.
