Find a number $n \neq 2017$ such that $\phi(n) = \phi(2017)$, as above. I know the formula for a general $\phi$ function, but I cannot see how this is helpful here. Any help would be appreciated!
Hint: $2017$ is a prime number.
Solution: Fortunately, the totient function $\phi$ has the nice property that $\phi(p) = \phi(2p)$, where $p$ is prime. This is because $2p$ is relatively prime to all odd numbers less than it except $p$. There are $p$ even numbers less than $2p$, and discounting the fact that $\gcd(p, 2p) \neq 1$, we see that $$\phi(2p) = p - 1 = \phi(p)$$ You could have alternatively seen this from the multiplicative property of $\phi$, namely that $\phi(mn) = \phi(m)\phi(n)$. Hence, $\phi(4034) = \phi(2017)$, because $\phi(2) = 1$.