Phi is 8 has only 5 solutions

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Theorem

$\phi$ denotes the Euler $\phi$-function.

There are $5$ numbers $n$ with the property that $\map \phi n = 8$, and they are $15$, $16$, $20$, $24$ and $30$.


Proof

Suppose $\phi(n)=8$.

By Fundamental Theorem of Arithmetic, $n$ has a prime factorization: $$n=\prod_{i=1}^k p_i^{\alpha_i}$$

where $p_i$ are all distinct primes. Since $\phi(1)=1\neq8$, we assume $k>0$. Say that $n$ has $k$ distinct prime factors, so that each $\alpha_i > 0$.

By Euler Phi Function is Multiplicative, $$\phi(n) = 8 =\prod_{i=1}^k \phi(p_i^{\alpha_i})$$

By Euler Phi Function of Prime Power, $$8 = \prod_{i=1}^k (p_i-1)p_i^{\alpha_i-1}$$

Because each $p_i-1$ is a factor of $8$, it follows that $p_i-1\leq 8$. Thus, each $p_i\leq 9$. The only primes less than $9$ are $2$, $3$, $5$, and $7$. Further, $7-1$ is not a factor of $8$, so $7$ is not a prime factor of $n$.

Thus, $n$ is of the form $2^{a}3^{b}5^{c}$, for nonnegative integers $a, b, c$.

By Euler Phi Function is Multiplicative and Euler Phi Function of Prime Power, if $b$ or $c$ are greater than $1$, then $\phi(n)$ is divisible by $3$ or $5$, respectively. This is not the case, so $3$ and $5$ each divide $n$ at most once.

This splits the analysis into four cases. Because $\phi(n)\leq n$ for all $n\in\mathbb{N}$, $n$ cannot be $1$, $3$, or $5$. Thus, in the first three cases, $n$ must have at least one factor of $2$.

Case 1:$b=c=0$

Now $n=2^a$, so $\phi(n)=8=(2-1)2^{a-1}=2^{a-1}\Rightarrow a=4$. This gives one solution of $n=16$.

Case 2: $b=1$ and $c=0$

$n=2^a\cdot 3$, so $\phi(n)=8=(2-1)2^{a-1}\cdot(3-1)\Rightarrow 8=2^a\Rightarrow a=3$. This gives one solution of $n=24$.

Case 3: $b=0$ and $c=1$

$n=2^a\cdot 5$, so $\phi(n)=8=(2-1)2^{a-1}\cdot (5-1)\Rightarrow 8=2^{a-1}\cdot 4\Rightarrow a=2$.This gives one solution of $n=20$.

Case 4: $b=c=1$

If $n$ is divisible by $2$:

$n=2^a\cdot 3\cdot 5$, so $\phi(n)=8=(2-1)2^{a-1}\cdot(3-1)\cdot(5-1)\Rightarrow 8=2^{a-1}\cdot 8\Rightarrow a=1$. This gives one solution of $n=30$.

If $n$ is not divisible by $2$, then $n=3\cdot 5$, and so $n=15$ is the fifth and final solution.


$\blacksquare$

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