Exercise 2.3.39 ($\phi(x) = n$ has only a finite number of solution)

Proof. Let $x>1$ be a solution of $\varphi (x)=n$. Write $x=p_1^{a_1}\cdots p_k^{a_k},a_i>0$ the decomposition of $x$ in prime factors. Then

Therefore $p_i-1$ is a divisor of $n$. Since the set of divisors of $n$ is finite, there are only finitely many possible $p_i$.

Write $n=q_1^{b_1}\cdots q_l^{b_l}$. If $a_i>1$, then $p_i^{a_i-1}\mid n$, so $p_i=q_j$ for some index $j$, and $a_i-1\le b_j$, so $0\le a_i\le b_j+1$, where $b_j$ is fixed, because $n$ is fixed. Thus there are only finitely possible exponents $a_i$.

This proves that the equation $\varphi (x)=n$ has only a finite number of solution. □