Is there an integer $N>0$ such that $\varphi(n) = N$ has infinitely many solutions?

Question

Let $\varphi: \mathbb{N} \to \mathbb{N}$ be the totient function.

Is there an integer $N > 0$ such that there are infinitely many integers $n > 0$ such that $$\varphi(n) = N?$$

Answer 1

No. Let $M\in\mathbb{Z}^{+}$ and let $p$ be the smallest prime such that $p> M+1$. Let $n\in\mathbb{Z}$ such that $\varphi(n)=M$. Now, suppose $q\ge p$ is a prime factor of $n$. Then, write $n=q^k\cdot m$, fro some $k\ge 1$ and $q\nmid m$. Then, we get $$\varphi(n)=\varphi(q^k\cdot m)=\varphi(q^k)\varphi(m)\ge q-1\ge p-1>M,$$ contradiction. Thus, no prime divisor of $n$ can be greater than $M+1$.

Thus, the prime divisors of $n$ make up a finite set, say $\{p_1,\ldots,p_m\}$. Now, write $n=p_1^{e_1}\cdots p_m^{e_m}$, for $e_i>0$. Then,

$$\varphi(n)=\varphi(p_1^{e_1})\cdots\varphi(p_m^{e_m})=\prod_{i=1}^{m}p_i^{e_i-1}(p_i-1).$$ Now, note that for each prime $p_i$, we have $\varphi(n)\ge p_i^{e_i-1}(p_i-1)>M$ for a sufficiently large choice of $e_i$. Thus, for each $p_i$ there are only finitely many valid choices for the $e_i$, and it follows that the set of all $n$ s.t. $\varphi(n)=M$ is finite.

Answer 2

No.

We use the fact that if $n = p_1^{e_1} \cdots p_k^{e_k}$, with the $p_i$ different primes and the $e_i$ positive, we have $\phi(n) = p_1^{e_1-1}(p_1-1) \cdots p_k^{e_k-1}(p_k-1)$.

Now consider an integer $N>0$ and suppose that $\phi(n) = N$. Then $n$ cannot be divisible by primes larger than $N+1$, since if $q > N+1$ divides $n$, we have $N+1 \leq q-1 \mid \phi(n) = N$, contradiction. The set of primes that may divide $n$, therefore, is finite. Let $p$ be such a prime. Then if $p^k \mid n$ we have $p^{k-1} \mid N$, which leaves a finite number of possibilities for $k$. Thus the number of primes that may divide $n$ is bounded and the exponent of each prime is bounded as well, thus there are at most finitely many $n$ for which $\phi(n) = N$.

Answer 3

No.

Theorem. If $n \geq 2^{2m}$ for positive integer $m$, then it follows that $\varphi(n) > m$.

Proof. Let $n=2^k\cdot r$ with an odd $r$. Then we have $\varphi(n) = \varphi(2^k) \varphi(r)$ by multiplicativity.

Now consider 2 cases:

1. If $k \leq m$, then $r = \frac{n}{2^k} \geq \frac{2^{2m}}{2^k} \geq 2^m$. Because $r$ is odd it means that every number $2^0, 2^1, 2^2, 2^3, \dots, 2^m$ is coprime with $r$ (there is at least $m+1$ such numbers). It now follows by definition of Euler's totient function that $\varphi(r) \geq m+1 > m$.

2. If $k > m$, then $\varphi(2^k) = 2^{k-1} \geq 2^m > m$.

In both cases we have that either $\varphi(2^k) > m$ or $\varphi(r) > m$. Because $\varphi(n) = \varphi(2^k) \varphi(r)$, we also have that $\varphi(n) > m$. $\blacksquare$

So for every $N$ by our theorem we have that when $n \geq 2^{2N},$ then it is impossible to have $\varphi(n) = N.$