Finite or Infinite n in $φ(n)$ Are there finitely or infinitely many integers n for which $φ(n) = 1000$? I think that there are finite, but I don't know how to prove it.
 A: Finitely many - $11$ in fact - the largest of which is $3750$. Here they all are:
$$\{1111,1255,1375,1875,2008,2222,2500,2510,2750,3012,3750\}$$
Taking $r = \dfrac{\phi(x)}{x}$, we know that $r = \prod \frac{p_i-1}{p_i}$ for each distinct prime power $p_i$ of $x$. So a new minimum $r$ is achieved for each primorial $x$, and the step up to the next primorial is far larger than the reduction of the ratio - $x$ will increase by a factor of $p_{n+1}$ (eg. say $13$) while minimum $r$ stays constant then only decreases by $\frac{p_{n+1}-1}{p_{n+1}}$ (say $\frac {12}{13}$) at that new primorial.
Looking at the $1000$ region, minimum $r$ is $\frac {1\times 2 \times 4 \times 6 \times 10}{2\times 3 \times 5 \times 7 \times 11} \approx 0.2078 $ (achieved at $2310$) so we definitely do not need to search past $x=\frac{1000}{0.2078}<5000$ for values where $\phi(x)=1000$. Just as a simple limiting argument.
Here's values of $\phi(x)$ up to $9000$ for illustrative purposes:

A: As Eric Naslund proved in this answer

For all $n\geq 3$ we have $$\phi(n)\geq \frac{n}{e^{\gamma}\log \log n}+O\left(\frac{n}{(\log \log n)^2}\right),$$ where $\gamma$ is the Euler-Mascheroni constant, and the above holds with equality infinitely often.

Further, observe that $n / \log \log n$ is increasing for every $n \geq 6$. In particular, we can find constants $N,c > 0$ such that
$$
\phi(n) \geq \frac{n}{e^{\gamma}\log \log n} - c \, \frac{n}{(\log \log n)^2} > 1000
$$
for every $n \geq N$.
This, in turn, implies that the positive integers $n$ for which $\phi(n) \leq 1000$ is bounded above by $N$, so there can only be finitely many of those.
A: For all $p \ge 3$, $1-\tfrac1p \ge p^{-1/2}$ so $\phi(p^r) > \sqrt{p^r}$.  Therefore $\phi(n) \ge \sqrt{n}$ for odd $n$, and accounting for even $n$, $\phi(n) \ge \tfrac12 \sqrt{n}$.  This easily gives $\phi(n) > 1000$ for all $n > 4000000$.
