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I want to prove the following:

Let $n \in \mathbb{N}$. Then, if

$$2\varphi(n) + 2 = n$$

holds, there exists an odd prime $p$ such that $n=2p$.

My guess is that one can use the multiplicative property of the totient function. For coprime numbers $m,n$

$$\varphi(m * n) = \varphi(m) \varphi(n)$$

In our case: $\varphi(2p) = \varphi(2) \varphi(p) = \varphi(p)$

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  • $\begingroup$ Would it help to factor out a $2$? $\endgroup$ – Mike Aug 26 '16 at 9:39
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The LHS is divisible by $2$, therefore RHS is divisible by $2$. Assume $n=2k$ with $k\in{\mathbb{Z}}$. In that case the equation becomes $$ 2\varphi(2k)+2=2k $$ which is equivalent to $$ \varphi(2k)+1=k. $$ If $k$ is odd, this yields $\varphi(k)=k-1$ by multiplicative property and therefore $k$ is prime and we are done. We like to show that this is the only possibility. Assume $k$ is even, write $k=2^m l$ with maximal $m$. Therefore $$ \varphi(2^{m+1}l)+1=2^ml. $$ Again using the multiplicative property of $\varphi$ this gives us $$ (2^{m+1}-2^m)\varphi(l)+1=2^ml $$ Here the RHS is divisible by $2$, but the LHS is not, which is a contradiction.

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