Euler, A.D. 2017

In general, $\phi(p_1^{e_1}\cdot p_2^{e_2}\cdots) = (p_1-1)p_1^{e_1-1}\cdot (p_2-1)p_2^{e_2-1}\cdots.$ Since $2017$ is a prime number, $2017|\phi(m)$ implies that either

• $p-1$ is a multiple of 2017, for some prime factor $p|m$ ;

• or $m$ contains at least two factors 2017.

Since we are looking for a smallest value, and the second option requires $n > m \geq 2017^2 > 4\:000\:000$ , we explore the first option, leaving the second for if that does not give any smaller result.

We need a prime number $p$ such that $p-1$ is a multiple of 2017. The smallest multiple for which this works is $p = 4\cdot 2017 +1 = 8069$ . Thus, $\phi(8069) = 8068 = 4\cdot 2017.$ Next, we find $n$ such that $\phi(n) = 8069$ . Following a similar reasoning, we find a prime number $q$ such that $q-1$ is a multiple of 8069. The smallest value that works is $q = 2\cdot 8069 +1 = 16\:139$ .

Thus, $\phi(16139) = 16138 = 2\cdot 8069,$ and combining everything we get $\phi(\phi(16139)) = \phi(2\cdot 8069) = \phi(2)\cdot \phi(8069) = 1\cdot 8068 = 2^2\cdot 2017,$ as desired.