Can I go further beyond the equation $\phi(n)=2d,$ where $d$ is odd greater than 1?

Question

Being inspired by the post and knowing that $\phi(n)$ must be even, I started to investigate the conditions which guarantee the existence of solutions of the equation $$\phi(n)=2d, \textrm{ where } d \textrm{ is an odd integer greater than 1.} \tag*{(*)} $$ I found a necessary and sufficient condition for the existence of the solution of the Totient equation.

$$\boxed{ \quad \textrm{ (*) has two solutions } p^\alpha \textrm{ and } 2p^{\alpha} \\\Leftrightarrow d=\frac{p-1}{2}p^{\alpha-1} \textrm{ for some natural number }\alpha \textrm{ and Gaussian prime } p.} $$

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Now I am going to prove it. First of all, $d$ is odd. For any $n$, we can decompose it into a product of its prime factors as

$$n=2^k \prod_{j=1}^m p_j^{\alpha_j}, \textrm{ where } p_j \textrm{ are disticnt odd primes and } k\geq 0 \tag*{} $$ By the properties of Euler’s Totient function: $ \phi(n)=n \prod_{p|n} \left(1-\frac{1}{p}\right)$ , we have $$ \phi\left(2^k\right) \prod_{j=1}^m\left(p_j-1\right) p_j^{\alpha_j-1}=2 d $$ Since $p_j-1$ is even for every $j$, therefore $m=1$ and hence

$$ \begin{aligned} & \phi\left(2^k\right)(p-1) p^{\alpha-1}=2 d\\ \Rightarrow & \phi\left(2^k\right)=1 \text { and }(p-1) p^{\alpha-1}=2 d\\ \Rightarrow & k=0 \text { or } 1 \text { and } d=\frac{p-1}{2} \cdot p^{\alpha-1} \text { for some Gaussian prime } p. \\ \Rightarrow & p^{\alpha}\textrm{ and }2p^{\alpha}\textrm{ are two solutions of }(*) \end{aligned} $$

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We can now conclude that

$$\boxed{ \quad \textrm{ (*) has two solutions } p^\alpha \textrm{ and } 2p^{\alpha} \\\Leftrightarrow d=\frac{p-1}{2}p^{\alpha-1} \textrm{ for some natural number }\alpha \textrm{ and Gaussian prime } p.} $$ For examples,

$\phi(n)=42$ has solutions $n=43,49, 86 \textrm{ and } 98$.$(\textrm{ Since } d= \frac{43-1}{2}\cdot 43^{1-1}=\frac{7-1}{2}\cdot 7^{2-1} )$

$\phi(n)=78$ has solutions $n=79 \textrm{ and }158 $.

$\phi(n)=14$ has no solutions.

My Question: Can I go further beyond the equation $\phi(n)=2d,$ where $d$ is odd?

Advices and alternative methods are highly appreciated?

Answer 1

After asking the question on the Totient equation $\phi(n)=2d$, where $d$ is an odd integer greater than $1$. I wanna try to raise the power of $2$ by $1$, $$\phi(n)=2^2d$$

Method 1

Since the number of prime factors of $n$ is bounded above by $2$, therefore there are only four possible forms for $n$, $$n= p^{\alpha} ,2p^{\alpha} ,p_1^{\alpha_1}p_2^{\alpha_2}\ \textrm{ or }2p_1^{\alpha_1}p_2^{\alpha_2} , $$

Considering their $\phi$ on both sides gives $$4d= (p-1)p^{\alpha-1} \textrm{ for some }p\equiv 1 \pmod 4 \\\textrm{ or } (p_1-1)p_1^{\alpha_1-1} (p_2-1)p_2^{\alpha_2-1} \textrm{ for some Gaussian primes }p_1 \textrm{ and }p_2\\\Rightarrow d=\frac{p-1}{4} p^{\alpha -1}\textrm{ for some prime }p\equiv 1 \pmod 4 \cdots (1) \\ \textrm{ or } d=\frac{(p_1-1)(p_2-1)}{4}p_1^{\alpha_1-1} p_2^{\alpha_2-1} \textrm{ for some Gaussian primes }p_1 \textrm{ and }p_2 \cdots (2) $$

Therefore we can conclude that :

$$\boxed{ \textrm{ A. }d \textrm{ satisfies }(1) \Rightarrow n= p^{\alpha} \textrm{ or } 2p^{\alpha}\quad \textrm{ B. }d \textrm{ satisfies }(2) \Rightarrow n= p_1^{\alpha_1} p_2^{\alpha_2} \textrm{ or } 2 p_1^{\alpha_1} p_2^{\alpha_2}}$$

for some prime $p\equiv 1 \pmod 4$ and Gaussian primes $p_1$ and $p_2.$

For example, when $\phi(n)=4\times 39=156$,

$$d=3\times 13=\frac{157-1}{4} \Rightarrow n=157 \text { or } 314 \\ \textrm{ and } d=\frac{13-1}{4} \cdot 13^{2-1} \Rightarrow n=13^2 \textrm{ or }2 \cdot 13^2=169 \textrm{ or }338$$ Using this method, we start with the prime factorisation of $d$ to catch the prime factors $p_i$’s of $n$.

Unfortunately, there are three solutions $237, 316 \textrm{ and } 474$ are left behind by this Method 1. We need another method.

Method 2.

We shall make use of the solutions $79$ and $158$ of $\phi(n)=78$, where $d$ is an odd integer greater than $1$. Using the property of Euler’s function $\phi(a\times b)=\phi(a)\times \phi(b)$, we have $$\phi(3\times 79)=\phi(3)\times \phi(79)=2\times 78=156\\ \phi(4\times 79)=\phi(4)\times \phi(79)=2\times 78=156\\ \phi(6\times 79)=\phi(2)\times \phi(79)=2\times 78=156$$ Therefore 3 more solutions are $237, 316 \textrm{ and }474.$ For further using this method to solve $\phi(n)=2^td$, we need to solve $\phi(n)=2^t$ Combining 2 methods, the full set solutions of $\phi(n)=156$ is $$\boxed{S=\{157,169,237,314,316,338,474\}} \tag*{} $$

**My Question:**May we raise the power of 2 further in order to solve the general Totient equation $\phi(n)=2^sd,\textrm{ where }d$ is odd and greater than 1.

Answer 2

You have an error in A..

(I) $d=ep^{\alpha}$ for some odd integer $e \ne 1$, prime $p$ and $e<p\Rightarrow n= p^{\alpha+1}$ or $2p^{\alpha+1}$

This is wrong. Taking $(e,p,\alpha)=(3,5,1)$, we have $\phi(25)=\phi(50)=20\not=30$.

You have to exclude $p$ satisfying $p\equiv 1\pmod 4$.

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As I commented, you have $d=\dfrac{p-1}{2}p^{\alpha-1}$ where we need $p\equiv 3\pmod 4$ since $d$ is odd.

We get $n=p^{\alpha},2p^{\alpha}$. Note that this works for $\alpha=1$.