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 $n$ ?
Best Answer
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 $\varphi(2q')=\varphi(q')$