Euler's Totient function $\forall n\ge3$, if $(\frac{\varphi(n)}{2}+1)\ \mid\ n\ $ then $\frac{\varphi(n)}{2}+1$ is prime

Solution 1:

It is not so difficult to see that your desired statement,

(1) $\forall n\ge3$, if $(\frac{\varphi(n)}{2}+1)\ \mid\ n\ $ then $(\frac{\varphi(n)}{2}+1)\in \Bbb P$

is true if the Lehmer Totient Conjecture would be true (I am not sure about only if). This means that the statement (1) is a simple consequence of the Lehmer Totient Conjecture (a very difficult conjecture).

Solution 2:

Now I understand that this is the generic version of expression (1):

$\forall\ n=pk\ ,\ p\in\Bbb N\ , $ if $\exists\ m\ /\ \varphi(m)=k\ $ , then $\frac{\varphi(pk)}{\varphi(k)}+1 = p \mid n$

So the property is true only if the denominator of the fraction is one of the possible values taken by the totient function (OEIS link here) and the numerator contains a prime and a $k$ whose value $\varphi(k)$ is exactly the value of the denominator.

For instance for the family $k\in \{5,8,10,12\}$ whose $\varphi(k) = 4$ , the valid $n$ are $n=kp$ and the valid expressions are $\frac{\varphi(p5)}{4}+1=p$,$\frac{\varphi(p8)}{4}+1=p$, $\frac{\varphi(p10)}{4}+1=p$,$\frac{\varphi(p12)}{4}+1=p \mid n$.

The sequence for the case $k=2$ is very easy to convert into code and check running through $n$ so it deserved a place in the OEIS database (link here).