It is known that given a positive integer $n$, the number of solutions of the equation $φ(x)=n$ is finite, where $φ$ denotes Euler's Phi function. So, I've made an attempt to prove this fact. Here's my attempt.
PROOF For any positive integer $n$ let $\mathscr L_n$ be the set of all $x \in \mathbb Z^+$such that $φ(x)=n$, and assume by way of contradiction that $\mathscr L_n$ is infinite for some positive integer $n$. Thus, there exists a sequence $(x_k)$ of all elements of $\mathscr L_n$. Now, since $φ(m) \ge \sqrt{\frac{m}{2}}$ for all $m \in \mathbb Z^+$, it is easy to see that $(x_k)$ is bounded from above by $2n^2$. Thus, $\mathscr L_n$ is bounded, a contradiction, because we can't have a subset of the naturals that is infinite and bounded, $ο.ε.δ.$
COMMENTS
- I know that there's another post on this here How to prove that this equation has finitely many solutions, but it was not helpful.
- It's also non-trivial that $φ(m) \ge \sqrt{\frac{m}{2}}$ for all $m \in \mathbb Z^+$, and there is a nice proof given here Prove that $\phi(n) \geq \sqrt{n}/2$.
EDIT
Due to an excellent observation by @user58697, I completely edited things.
