# 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?

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.
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.