# Euler's totient function and prime factorization

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: $\varphi(2p) = \varphi(2) \varphi(p) = \varphi(p)$

• Would it help to factor out a $2$? – Mike Aug 26 '16 at 9:39

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.