Euler's function $\phi$: Values such that $\phi(n)=8$, $\phi(n)=14$

Question

Let $\phi(n) $ be Euler's Totient Function

Let us consider $$ |\{ n \in \mathbb{N} : \phi (n) = 8 \} | = 5, $$ and $$ |\{ n \in \mathbb{N} : \phi (n) = 14 \} | = 0. $$

How would I go about proving this?

Answer 1

If there are $r\ge1$ odd primes that divide $n$, then $2^r$ divides $\phi(n)$ (*). And if $2^t$ ($t\ge1$) divides $n$ then $2^{t-1}$ divides $\phi(n)$.

Therefore, if $\phi(n)=14$, then $n$ is the power of an odd prime ($n=p^s$) or its double ($n=2p^s$). Any case, $\phi(n)=p^{s-1}(p-1)$ so we have two possibilities:

• If $s>1$ then $p^{s-1}$ divides $14$, that is, $p=7$ and $s=2$, but $\phi(7^2)=42$.
• If $s=1$ then $p=n$ and $\phi(n)=n-1$, but $15$ is not prime.

Thus, there is no $n$ such that $\phi(n)=14$.

Can you now try with $8$? It's not very different.

Proof of (*): Let $p_1,\ldots,p_r$ be the odd prime factors of $n$. Then $$n=2^k\prod_{j=1}^r p_j^{\alpha_j}$$ for some $k\ge 0$ and $$\phi(n)=\phi(2^k)\prod_{j=1}^r(p_j-1)p_j^{\alpha_j-1}$$

Sinnce every $p_j-1$ is even, $2^r$ divides $\phi(n)$.

Answer 2

In my solution, I would use properties: $\phi(a\times b)=\phi(a)\times \phi(b)$ for any $(a, b)=1$ and $\phi(p^{\alpha})=p^{\alpha-1}(p-1)$ First of all, let’s first solve the equation $ \displaystyle \phi(n)=8.$

A. When $n$ is odd, let the prime factorisation of $n$ be $ \displaystyle \displaystyle n=\prod_{j=1}^m p_j^{\alpha_j} \quad \textrm{ where } p_j \textrm{ are ascending odd primes }, \tag*{} $ $\quad $ then use the properties, we have $\displaystyle \begin{aligned}& \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}=2^3 \\\Rightarrow \quad & \alpha_j=1 \quad \textrm{ and } \quad \prod_{j=1}^m\left(p_j-1\right)=2^3 \\\Rightarrow \quad & p_1=3 \textrm{ and } p_2=5 \\\Rightarrow \quad & n=3 \times 5=15\end{aligned}\tag*{} $

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B. When $n$ is a power of $2$, i.e. $n=2^m,$ $\displaystyle 2^{m-1}=\phi(m)=8 \Rightarrow m=4 \textrm{ and }n=16.\tag*{} $

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C. When $n$ is even with odd factor(s), its prime factorisation is $\displaystyle N=2^k \prod_{j=1}^m p_j^{\alpha_j}, \textrm{ where } p_j \textrm{ are ascending odd primes and } k\geq 1 \tag*{} $ $\quad $ then $\displaystyle \begin{array}{ll} & \displaystyle 2^3=\phi(n)=\phi\left(2^k\right) \prod_{j=1}^m \phi\left(p_j^{\alpha_j}\right)=2^{k-1} \displaystyle \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}\\ \Rightarrow & 2^{4-k}= \displaystyle \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}\\\Rightarrow &\alpha_j=1 \quad \text { and } \quad p_j=2^{r_j}+1 \quad \forall j=1,2, \ldots m\\\Rightarrow & 2^{4-k}=\displaystyle \prod_{j=1}^m 2^{r_j }=2^{r_1+r_2+\cdots+r_m} \\\Rightarrow & r_1+r_2+\ldots+r_m=4-k \\ \end{array}\tag*{} $

(i) When $k=1$, $\displaystyle r_1+r_2+\ldots+r_m=3 \Rightarrow m=2,r_1=1 \textrm{ and }r_2=2 \Rightarrow n=2\times 3\times 5=30 \tag*{} $

(ii)When $k=2$, $\displaystyle r_1+r_2+\ldots+r_m=2 \Rightarrow m=1,r_1=2 \Rightarrow n=2^2\times 3=20 \tag*{} $

(iii)When $k=3$, $\displaystyle r_1+r_2+\ldots+r_m=1 \Rightarrow m=1,r_1=1\Rightarrow n=2^3\times 3=24 \tag*{} $

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Therefore we can conclude that there are five integers $\boxed{15, 16, 20, 24 \textrm{ and }30}$ satisfying $\phi(n)=8$.

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Now we are going to prove that $\phi(n)=14$ has no solutions. For any natural number $n$, its prime factorisation is $\displaystyle N=2^k \prod_{j=1}^m p_j^{\alpha_j}, \textrm{ where } p_j \textrm{ are ascending odd primes and } k\geq 0 \tag*{} $

Then $$ \phi\left(2^k\right) \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}=2 \times 7 \Rightarrow m=1, $$ for otherwise the product $\prod_{j=1}^m\left(p_j-1\right)$ is divisible by $4$ while the right expression is only divisible by $2$, which is a contradiction. $$ \phi\left(2^k\right)\left(p_1-1\right) p_1^{\alpha_1-1}=2\times 7 $$

If $k=0$ or $1$, then $\left(p_1-1\right) p_1^{\alpha_1-1}=2\times 7 \Rightarrow p_1=3 \textrm{ or }15,$ a contradiction.

If $k\geq 2$, then $2^{k-1}\left(p_1-1\right) p_1^{\alpha_1-1}=2\times 7 \Rightarrow 2^{k-2}\left(p_1-1\right) p_1^{\alpha_1-1}=7 \Rightarrow$ the left expression is even and the right is odd, a contradiction.