$p \geq 5$ prime. $2p+1$ not a prime. Then $\phi(n)=2p$ has no solution. 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. 
 A: 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 $p\neq 2$.
A: 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 $2.$
A: First note that the condition $2p+1$ is not prime is necessary to prove that $phi(n)=2p$ has no solution, since $\phi(2p+1)=2p$ if $2p+1$ is prime.
Next note that if $n$ has $k$ distinct odd prime divisors $q_1,\ldots,q_k$, then $2^k \mid \phi(n)$ since $(q_1-1) \cdots (q_k-1) \mid \phi(n)$. So for $\phi(n)=2p$ to hold, $n$ can have at most one odd prime divisor. The presence of $p$ is $\phi(n)$ ensures that $n$ can't be a power of $2$. So $n$ must have exactly one odd prime divisor. This already contributes $2$ to $\phi(n)=2p$, so the power of $2$ in $n$ is at most $1$.
Thus, $n=2^e q^{\alpha}$, where $e \in \{0,1\}$ and $\alpha \ge 1$. From
$$ 2p = \phi(n) = q^{\alpha-1} (q-1) $$
we have $p \mid q$ or $p \mid (q-1)$. In the latter case, $2p \mid (q-1)$, which can only happen when $q=2p+1$. Since $2p+1$ isn't prime by assumption, the first case applies, and $q=p$. Hence, $2p=p^{\alpha-1} (p-1)$, which is only possible if $\alpha=2$ and $p-1=2$. This case is ruled out because then $2p+1=7$ is prime, and also because we assumed $p \ge 5$.
Now suppose we wish to find all $n$ such that $\phi(n)=2p$, where $2p+1$ is prime. As before, we have $2p=q^{\alpha-1} (q-1)$, and there is no solution when $p \mid q$.
So the only possibility is $p \mid (q-1)$, and in this case we must have $q=2p+1$, as seen above. We must also have $\alpha=1$. Together with $e \in \{0,1\}$, we have
$$ \phi(n) = 2p \Longleftrightarrow n = 2p+1 \:\text{or}\: 2(2p+1). $$
A: Noting that when $n $ satisfies the equation $\phi(n)=2p$, where $p$ is prime, $$n=2^k q^{\alpha} \Rightarrow \phi(2^k)(q-1)q^{\alpha-1}= (q-1)q^{\alpha-1} =2p , \textrm{  where } q \textrm{  is an odd prime.} $$
Knowing that $2|(q-1)$ for any odd prime $q$, we have
$$\qquad\qquad ^{\prime} k=0 \textrm{  or }1^{\prime}
\textrm{  and }’q=3 \textrm{  and }\alpha = 2’\textrm{  or }’q-1=2p \textrm{ and }\alpha=1’\\ \Leftrightarrow n=9,18, 2p+1 \textrm{ or } 2(2p+1) \textrm{ (accepted  only if }2p+1 \textrm{  is prime.)}$$
More precisely, I would like to restate it as :
If $p$ is a prime greater than $3$,
$$\boxed{\textrm{  The equation }\phi(n)=2p \textrm{  has solutions }2p+1 \textrm{ and } 2(2p+1)\Leftrightarrow p  \textrm{  is a Germain prime.   } }$$
Since $p=3$ is also a Germain prime, therefore we have four solutions $n=9,18,7 \textrm{  or }14$ for the equation $\phi(n)=6.$
