how to prove that there don't exist any $n\in \Bbb N$ such that $\phi (n)=14$? How to prove that there don't exist any $n\in \Bbb N$ such that $\phi (n)=14$ ?
We know that
$\phi (n)= {p_1}^{\alpha_1}{p_2}^{\alpha_2}...{p_n}^{\alpha_n}(1-\frac{1}{p_1})...(1-\frac{1}{p_n}) $
if $n=3$ then $\phi (n)=2$ also we have $\phi (9)=6$ and $ \phi(n_1)\phi(n_2)=\phi(n_1n_2)$.
So the  problem turns out to be :
Does there exist $n \in \Bbb N$ such that $\phi(n)=7$ ?
 A: You have reduced the problem to showing that there is no $n$ satisfying $$\phi(n)=7$$.
But this is very simple, as $\phi(n)$ must be even for $n\ge3$, since the expression for $\phi(n)$ contains $p-1$. So the result follows.
A: Recall that $$\varphi(ab)=\varphi(a)\varphi(b)$$ for $a,b$ coprime and that $$\varphi(p^r) = (p-1)p^{r-1}$$
Notice that $\varphi(n) = 2 \cdot 7$ is a product of two distinct primes, so $n$ is either a prime power or a product of two prime powers. Therefore we are reduced to a Diophantine problem


*

*$2 \cdot 7 = (p-1) p^{r-1}$ is impossible.

*$2 \cdot 7 = (p-1) (q-1) p^{r-1} q^{k-1}$ is impossible.

A: We have $$\tag1\phi(n)=p_1^{\alpha_1-1}p_2^{\alpha_2-1}\cdots p_k^{\alpha_k-1}(p_1-1)(p_2-1)\cdots(p_k-1)$$ if $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot p_k^{\alpha_k}$ with primes $p_1<p_2<\ldots <p_k$ and $\alpha_i\in\mathbb N$.
If $\phi(n)=14$ then one of the factors on the right must be divisible by $7$, that is we have $p_i=7$ ocurring with $\alpha_i\ge 2$; or there is some $p_i\equiv 1\pmod 7$. In the first case, we obtain that $\phi(n)$ is a multiple of $7^1\cdot (7-1)=42$. In the latter case $\phi(n)$ is a multiple of $p_i-1$, where $p_i$ is a prime $\equiv 1\pmod 7$, hence $p_i\ge 29$. In bothe cases, we see $\phi(n)> 14$, contradiction.
