Is the equation $x = \phi(n), x=2k, n,k \in \mathbb{Z}$, where $\phi(n)$ is the Euler totient function, solvable for all evens? I was just getting my hands dirty solving some equations of the form $x=\phi(n)$ where $\phi(n)$ is Euler totient function. I know that $\phi(n)$ is even for $n\geq 3$. However, I am wondering that:

Can all even integers $x$ be expressed as $\phi(n)$ where $n$ is an integer?

I start doing some research, the first even number appears to hove no solution is $14$. According to this page, there is no solution to the equation $14=\phi(n)$ for $n\leq 500$. But does it have a solution for $n>500$?
Moreover, what can you say about the above statement?
 A: 14 is not $\varphi(n)$ for any $n$. Such numbers are called nontotients, there are infinitely many of them and they are given by OEIS A005277.
To show the infinitude of nontotients, consider the primes $p=10n+7$, which are themselves infinite by Dirichlet's theorem on arithmetic progressions. Then $2p+1$ is composite, being divisible by 5, and by a comment by Firoozbakht on the OEIS page $2p$ is a nontotient. In particular, 14 is the first nontotient corresponding to a $10n+7$ prime.
A: To see that $\varphi(n)=14$ has no solution, consider the possible prime factorization of $n$.  If a prime $p>15$ divided $n$, then $p-1$ would divide $14$, a contradiction.  So $n$ can only be divisible by $2,3,5,7,11,13$.  Easy to eliminate $7,11,13$, so $n=2^a3^b5^c$.   But $7$ clearly doesn't divide $\varphi(2^a3^b5^c)$, and we are done.
To see that there are infinitely many $m$ for which $\varphi(n)=m$ has no solution, consider $m$ of the form $2p$ where $p$ is an odd prime, such that $2p+1$ is not prime (it's enough that $p\equiv 1 \pmod 3$, say).  Suppose that $\varphi(n)=2p$, and let $q$ be a prime dividing $n$. Now $(q-1)\,|\,2p\implies (q-1)\in \{1,2,p,2p\}$  Thus $q\in \{2,3\}$.  Easy to see that this is impossible.
A: Now we are going to prove that $\phi(n)=14$ has no solutions.
For any natural number $n$, its prime factorisation is
$\displaystyle n=2^k \prod_{j=1}^m p_j^{\alpha_j}, \textrm{ where } p_j \textrm{ are ascending odd primes and } k\geq 0 \tag*{} $
Then $$
\phi\left(2^k\right) \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}=2 \times 7 \Rightarrow m=1,
$$
for otherwise the product $\prod_{j=1}^m\left(p_j-1\right)$  is divisible by $4$  while the right expression is only divisible by $2$, which is a contradiction.
$$
\phi\left(2^k\right)\left(p-1\right) p_1^{\alpha-1}=2\times 7
$$
Since $2|(p-1)$, therefore $\phi\left(2^k\right)=1$ and $\left(p-1\right) p^{\alpha-1}=2\times 7 \Rightarrow p=3 \textrm{ or }15,$ a contradiction.
Therefore we can conclude that there no solution for $\phi(n)=14.$
