$p \geq 5$ prime. $2p+1$ not a prime. Then $\phi(n)=2p$ has no solution.

Question

I am studying for my number theory final by doing past exams. The question is

Let $p \geq 5$ be a prime. Prove that if $2p+1$ not a prime. Then $\phi(n)=2p$ has no solution. ($\phi$ is Euler Totient function). In the case where $2p+1$ is prime, find all solutions to $\phi(n)=2p$.

I managed to 'prove' the first part, however without using the $2p+1$ not prime assumption. I spot if $2p+1$ is prime then $\phi(2p+1)=2p$, assuming the first part this shows that $\phi(n)=2p$ has solutions $\iff 2p+1$ is prime. But how to find all solutions? So clearly the $2p+1$ not prime assumption is needed. So my proof is wrong but I can't see where it fails (edit: as pointed out I just missed a case, the proof itself is not wrong). Can someone show me where it fails, and or provide an alternative. My proof:

Let $n=2^a\prod_{i=1}^tp_i^{\alpha_i}$ be the prime factorisation of $n$ with $a\geq 0$, and $p_i$ odd primes. Assume that $\phi(n)=2p$. Know that $$\phi(n)=\phi(2^a)\prod_{i=1}^t\phi(p_i^{\alpha_i})=2p.$$ If $t\geq 2$, then $4 \mid 2p$, since $\phi(p_i^{\alpha_i})$ is even. Hence we must have $t < 2$. If $t=0$ then $\phi(n)=2^{a-1} \neq 2p$. So we must have that $t=1$ and so we have $$n=2^ap_1^{\alpha_1}.$$ If $a>2$ then $ 4 \mid \phi(2^a) \mid 2p$, which is a contradiction. So we must have $a \leq 2$. If $a=0$ or $a=1$, then $\phi(2^a)=1$ and we have $\phi(n)=p_1^{\alpha_1-1}(p_1-1)=2p$. Hence $p_1-1=2$ and $p=p_1^{\alpha_1-1}$, which leads to $p_1=p=3$, which is a contradiction.

Hence we must have $a=2$, i.e $n=4p_1^{\alpha_1}$. Then $\phi(n)=2p_1^{\alpha_1-1}(p_1-1)=2p$. Which implies $p_1^{\alpha_1-1}(p_1-1)=p$. So either $p_1-1=p$ , in which case they are consecutive primes, contradiction as $p\geq 5$. Or $p_1^{\alpha_1-1}=p$ and $p_1-1=1$, which leads to $p_1=p=2$, also a contradiction. So all cases lead to a contradiction.

Answer 1

In the case that $a=0,1$, you have used that $2p+1$ is not prime when you said that $\phi(n) = p_1^{\alpha_1 -1}(p_1 -1) = 2p$ gives $p_1-1 = 2$ and $p = p_1^{\alpha_1 - 1}$. Indeed we could instead have $p_1-1 = 2p$, and $\alpha_1 = 1$. But this is ruled out since then $p_1 = 2p+1$ is not prime.

You should also note that we could, a priori, have $p_1 - 1=p$, but then $p_1$ is not prime since $p\neq 2$.

Answer 2

Suppose $\phi(n)=2p.$ Then $n\ge \phi(n)>1$ so $n>1$ so $n$ has at least one prime divisor $q$, and $(q-1) | \phi(n)=2p.$ But the only divisors of $2p$ are $1,2, p,2p$ because $p$ is prime.

So $q-1\in\{1,2,p,2p\}$ so $q\in \{2,3,p+1,2p+1\}.$ But $p+1$ is even and $>2,$ hence not prime, and $2p+1$ is not prime. So $q\in \{2,3\}.$

And $3^2\not| \,n, $ else $3|\phi(n)=2p,$ implying $3\in \{1,2,p,2p\}$ with $p\ge 5,$ which is impossible.

So $n=2^A$ or $n=2^A\cdot 3$ for some $A\in \Bbb N,$ or $n=3.$ But for each of these possible values, $\phi(n)$ is a power of $2.$