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

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

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \map \phi {33 \, 817 \, 088}
      | r = 16 \, 842 \, 752
      | c = {{EulerPhiLink|33,817,088|33 \, 817 \, 088}}
}}
{{eqn | r = 2^{16} \times 257
      | c = 
}}
{{end-eqn}}
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, we have:
:$p - 1 \divides p^{k - 1} \paren{p - 1} =  \map \phi {p^k} \divides \map \phi x$
where $k$ is the highest power of $p$ dividing $x$.
Therefore, by considering divisors of $\map \phi x$, 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 $1 \le j \le 16$:
{{begin-eqn}}
{{eqn | q = j = 1
      | l = 2^1 \times 257 + 1
      | r = 515
      | rr= = 5 \times 103
}}
{{eqn | q = j = 2
      | l = 2^2 \times 257 + 1
      | r = 1029
      | rr= = 3 \times 7^3
}}
{{eqn | q = j = 3
      | l = 2^3 \times 257 + 1
      | r = 2057
      | rr= = 11^2 \times 17
}}
{{eqn | q = j = 4
      | l = 2^4 \times 257 + 1
      | r = 4113
      | rr= = 3^2 \times 457
}}
{{eqn | q = j = 5
      | l = 2^5 \times 257 + 1
      | r = 8225
      | rr= = 5^2 \times 7 \times 47
}}
{{eqn | q = j = 6
      | l = 2^6 \times 257 + 1
      | r = 16 \, 449
      | rr= = 3 \times 5483
}}
{{eqn | q = j = 7
      | l = 2^7 \times 257 + 1
      | r = 32 \, 897
      | rr= = 67 \times 491
}}
{{eqn | q = j = 8
      | l = 2^8 \times 257 + 1
      | r = 65 \, 793
      | rr= = 3 \times 7 \times 13 \times 241
}}
{{eqn | q = j = 9
      | l = 2^9 \times 257 + 1
      | r = 131 \, 585
      | rr= = 5 \times 26 \, 317
}}
{{eqn | q = j = 10
      | l = 2^{10} \times 257 + 1
      | r = 263 \, 169
      | rr= = 3^6 \times 19^2
}}
{{eqn | q = j = 11
      | l = 2^{11} \times 257 + 1
      | r = 526 \, 337
      | rr= = 7 \times 17 \times 4423
}}
{{eqn | q = j = 12
      | l = 2^{12} \times 257 + 1
      | r = 1 \, 052 \, 673
      | rr= = 3 \times 350 \, 891
}}
{{eqn | q = j = 13
      | l = 2^{13} \times 257 + 1
      | r = 2 \, 105 \, 345
      | rr= = 5 \times 11 \times 101 \times 379
}}
{{eqn | q = j = 14
      | l = 2^{14} \times 257 + 1
      | r = 4 \, 210 \, 689
      | rr= =  3 \times 7 \times 43 \times 4663
}}
{{eqn | q = j = 15
      | l = 2^{15} \times 257 + 1
      | r = 8 \, 421 \, 377
      | rr= = 1123 \times 7499
}}
{{eqn | q = j = 16
      | l = 2^{16} \times 257 + 1
      | r = 16 \, 842 \, 753
      | rr= = 3^2 \times 1 871417
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map \phi {257^2 \times 65 \, 537}
      | r = 257 \times 256 \times 2^{16}
}}
{{eqn | o = >
      | r = 2^{16} \times 257
}}
{{end-eqn}}
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 $1$.
{{AimForCont}} $x$ is odd, that is, $y = 0$.
We have that:
{{begin-eqn}}
{{eqn | l = \map \phi x
      | r = \map \phi {3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2}
}}
{{eqn | r = 2^{\epsilon_0} \times 4^{\epsilon_1} \times 16^{\epsilon_2} \times 256 \times 257
      | c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | r = 2^{\epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8} \times 257
      | c = 
}}
{{eqn | o = \le
      | r = 2^{1 + 2 + 4 + 8} \times 257
      | c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | o = <
      | r = 2^{16} \times 257
      | c = 
}}
{{eqn | r = \map \phi x
}}
{{end-eqn}}
which is a contradiction.
Thus there can be no odd integer $x$ satisfying $(1)$, that is, $y \ge 1$.
Note that:
{{begin-eqn}}
{{eqn | l = \map \phi {2^y \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2}
      | r = 2^{y - 1} \times 2^{\epsilon_0} \times 4^{\epsilon_1} \times 16^{\epsilon_2} \times 256 \times 257
      | c = as $\epsilon_i = 0$ or $1$
}}
{{eqn | r = 2^{y - 1 + \epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8} \times 257
      | c = 
}}
{{eqn | r = 2^{16} \times 257
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y - 1 + \epsilon_0 + 2 \epsilon_1 + 4 \epsilon_2 + 8
      | r = 16
}}
{{eqn | ll= \leadsto
      | l = y
      | r = 9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2
}}
{{end-eqn}}
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.
{{qed}}