Exercise 2.3.34 ($\phi(x) = 14$ has no solution)

Proof. If $x=p_1^{a_1}\cdots p_k^{a_k}$ is a solution of $\varphi (x)=14$, where $a_i>0$, then

Therefore $p_i-1\mid 14$ for every index $i$, so $p_i-1\in \{1,2,7,14\}$, where $p_i$ is prime, so $p_i=2$ or $p_i=3$. Therefore

If $b>1$, then $3\mid \varphi (n)=14$. This is false, so $b=0$ or $b=1$.

If $a>1$, then $2^{a-1}\mid 14=2\cdot 7$, thus $a-1\le 1$, $a\le 2$. So $0\le a\le 2$.

Therefore

so

But $\varphi (1)=\varphi (2)=1,\varphi (4)=\varphi (3)=\varphi (6)=2,$ and $\varphi (12)=4$. Therefore there is no solution of the equation $\varphi (x)=14$.

The integer $14$ is the first even integer such that $\varphi (x)=14$ has no solution. Indeed, for every even integer $n$ less than $14$, there is at least a solution of the equation $\varphi (x)=n$, as seen in the following array:

With similar methods, we see that the next even integer $n$ such that $\varphi (x)=n$ has no solution is $26$. □