Exercise 2.3.32 (Solve $\phi(x) = 24$.)

Proof. Let $x$ be a solution of $\varphi (x)=24$, and

its decomposition in prime numbers. Then

therefore $p_i-1\mid 24$ for every index $i$.

The divisors of $24$ are

Since $p_i-1$ is a divisor of $24$, and $p_i$ is prime,

Moreover, if $a_i\ge 2$, then $p_i$ and $p_i-1$ are divisors or $24$, thus $p_i=2$ or $p_i=3$.

If $p_i=3$, then $3^{a_i-1}\mid 24=2^3\cdot 3$, so $a_i-1\le 1,a_i\le 2$.

If $p_i=2$, then $2^{a_i-1}\mid 24=2^3\cdot 3$, so $a_i-1\le 3$, $a_i\le 4$. Note that $a_i=4$ is impossible, because there is at least another odd factor $q$, which gives another even factor $q-1$ of $\varphi (x)=24$.

To summarize,

$x$ is a divisor of $32760=2^3\cdot 3^2\cdot 5\cdot 7\cdot 13$. This gives $96$ numbers to check.

With Sage:

l = []
for a in range(4):
    for b in range(3):
        for c in range(2):
            for d in range(2):
                for e in range(2):
                    n = 2^a * 3^b * 5^c * 7^d * 13^e
                    if euler_phi(n) == 24:
                        l.append(n)
print(l)
              [35, 39, 45, 70, 78, 90, 52, 84, 56, 72]

There are 10 solutions of $\varphi (x)=24$, which are