Prove for any given positive integer $N$ there exist only finitely many integers n with $φ(n)=N$ [duplicate]

Question

Prove for any given positive integer $N$ there exist only finitely many integers $n$ with $φ(n)=N$, where $φ$ denotes Euler’s $φ$-function. Conclude in particular that $φ(n)$ tend to infinity as n tend to infinity.

How can I prove the above statement? Because Euler’s $φ$-function is bijection?

Answer 1

Let $N$ be a positive integer, $p$ be the least prime greater than $N+1$, and $n$ be any integer such that $\varphi(n) = N$. We now show that there are finitely many such $n$:

If $q \geq p$ is a prime divisor of $n$, then $n = q^{k}m$ for some $k, m \geq 1$ and $gcd(q^{k}, m) = 1$. Furthermore, $\varphi (n) = \varphi (q^{k}) \varphi (m) \geq q-1$ since $\varphi (m) \geq 1$ and $\varphi (q^{k}) = q^{k-1}(q - 1) \geq q - 1$. This leads us to a contradiction; for $q-1 \geq p-1 \geq N$. Thus, no prime divisor of $n$ is greater than $N+1$. In particular, the distinct prime divisors of $n$ belong to a finite set, say $\{p_1, p_2, .., p_m\}$.

Now, consider the prime decomposition of $n$, $p_1^{a_1}p_2^{a_2}...p_n^{a_n}$, then $\varphi(n) = \varphi(p_1^{a_1})\varphi(p_2^{a_2})...\varphi(p_m^{a_m})$. Thus, we have $\varphi(n) = \prod_{i = 1}^m p_i^{a_i-1}(p_i - 1)$. Note that, if $p^k$ is a factor of $n$, then $p^{k-1}$ must be a factor of $N$, which ultimately limits the possible values for $k$. Therefore, $n$ must be a product of a bounded set of primes, each raised to a bounded exponent; there are finitely many such $n$.

Finally, for each positive integer $N$, there is a largest integer $n$ with $\varphi(n) = N$. Thus, as $n$ approaches infinity, so does $\varphi(n)$.