Smallest Even Integer whose Euler Phi Value is not the Euler Phi Value of an Odd Integer

Theorem
The smallest even integer whose Euler $\phi$ value is shared by no odd integer is $33 \, 817 \, 088$.

Proof
We have:

Consider the equation:
 * $(1): \quad \map \phi x = 2^{16} \times 257$

Let $p$ be an odd prime factor of $x$.

Then as Euler Phi Function is Multiplicative, either:
 * $p - 1 = 2^j$ for some $j \in \Z$ such that $1 \le j \le 16$

which corresponds to the Fermat primes, or:
 * $p - 1 = 2^j \times 257$ for some $j \in \Z$ such that $1 \le j \le 16$

But $2^j \times 257 + 1$ is composite for $1 \le j \le 16$:

So the only possible odd prime factors of $x$ are the Fermat primes not exceeding $2^{16} + 1$:
 * $3, 5, 17, 257, 65 \, 537$

such that:
 * $257$ occurs with multiplicity $2$, since we need $257$ to be a factor of $\map \phi x$ but not $257^2$;
 * all other prime factors occurs with multiplicity $1$, since we do not want these primes to be factors of $\map \phi x$.

As we have:

it follows that $257$ and $65 \, 537$ cannot appear together.

We can now write:
 * $x = 2^y \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$

where each $\epsilon_i = 0$ or $1$.

$x$ is odd, that is, $y = 0$.

We have that:

which is a contradiction.

Thus there can be no odd integer $x$ satisfying $(1)$, that is, $y \ge 1$.

Note that:

Therefore even solutions to $(1)$ are of the form:
 * $x = 2^{9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2} \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$

Since each of:
 * $2^{-1} \times 3^1$
 * $2^{-2} \times 5^1$
 * $2^{-4} \times 17^1$

are greater than $1$, taking $\epsilon_i = 0$ for $i = 0, 1, 2$ would result in the smallest even integer satisfying $(1)$, that is:
 * $x = 2^9 \times 257^2 = 33 \, 817 \, 088$

It can be established by computer that $33 \, 817 \, 088$ is the smallest even integer whose Euler $\phi$ value is shared by no odd integer.