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