Can the smallest solution of $\varphi(k)=n$ be an even positive integer? Suppose, $n$ is a positive integer and the equation $$\varphi(k)=n$$ has a solution. Upto $n=20\ 000$, the smallest solution $k$ (if existent) is an odd positive integer.

Do positive integers $n$ exist, such that the smallest solution of $\varphi(k)=n$ is an even integer ? If yes, which is the smallest such $n$ ?

 A: I don't know which $n$ is smallest, but I know $n=2^{32}$ satisfies the condition (look up Fermat numbers if you want to figure out why for yourself).
Since $n$ is a power of 2, $k$ must be of the form $2^a*p_1*...p_n$ where $p_i$ are distinct odd primes that are equal to a power of 2 plus 1. As it happens, $2^1+1$, $2^2+1$, $2^4+1$, $2^8+1$ and $2^{16}+1$ are all prime (see Fermat numbers) but up to $2^{2048}+1$ no other such primes are known, so any solution of the equation $\varphi(k)=2^{32}$ must have $k$ divisible by 4 (since $\varphi(p_1*...p_n)$ can only get you to $2^{31}$ and in particular $k$ will be even.
Edit: I can't comment but the mistake in the other answer is where he implicitly uses near the end that $\varphi(2q')=2\varphi(q')$ but $q'$ is odd so in fact $\varphi(2q')=\varphi(q')$ 
A: Building on the ideas in the answer by anonymous, I have managed to construct a smaller example, namely $2^{16}\cdot257 = 16842752$.  As was shown with $2^{32}$, I'll show that if $\phi(k) = 2^{16}\cdot257$, then $k$ must be even.
Start by considering what prime power $p^a$ in the factorization of $k$ could yield the factor of $257$ in $\phi(k)$.  A priori, there are two possibilities:


*

*$a = 1$: In this case $\phi(p) = p - 1$ must be a multiple of $257$ and divisor of $2^{16}\cdot257$, so $p = 2^b\cdot257 + 1$ for some $b$, with $0\le b\le 16$.  It turns out that there is no such prime.  When $b$ is even, $2^b\cdot257 + 1$ is divisible by $3$; when $b\equiv1\pmod4$, it is divisible by $5$; when $b\equiv3\pmod8$, it is divisible by $17$; and when $b = 7$ or $15$, it is divisible by $67$ or $1123$, respectively.  (This fortunate numerical coincidence is why $257$ was chosen.)

*$a \gt 1$: In this case, $p$ divides $\phi(k)$, so $p$ is either $2$ or $257$.  But $\phi(2^a) = 2^{a-1}$ is never a multiple of $257$, so $p = 257$.  Moreover, if $a$ were greater than $2$, then $\phi(257^a)$ would be divisible by $257^2$.  Thus $a=2$, so $k$ must have a prime-power factor of $257^2$.
Since $\phi(257^2) = 2^8\cdot257$, that leaves a factor of $2^8$ in $\phi(k)$ still to be accounted for.  As in anonymous's argument, it must come from some combination of factors of $2$ and Fermat primes that divide $k$.  But since we've already used $257$, the remaining smaller Fermat primes cannot by themselves produce the additional eight factors of $2$ in $\phi(k)$.  Hence $k$ is even; in particular, the smallest possible $k$ is $2^9\cdot257^2$.
Having figured out that $16842752$ works, it is now easy find out that it has been discovered before.  In particular, it is the first entry in sequence A143510 in the On-Line Encyclopedia of Integer Sequences.  This sequence shows all nine known integers $n$ such that $\phi(k) = n$ only for even values of $k$.  ($2^{32}$ is the largest of the nine.)
This still does not establish that $16842752$ is the smallest value for the question asked here, but it seems likely.  The value is small enough that it could reasonably be verified computationally.
EDIT: I should have said that sequence A143510 contains the nine smallest integers $n$ such that $\phi(k) = n$ only for even values of $k$; additional larger values are known, but there may be others unknown in between.  In a note titled Numbers Like 16842752, T.D. Noe lists many larger values, as well as observing that each is a member of an infinite family obtained by multiplying by suitable powers of $2$.
