Euler's Phi function, elementary number theory Show that the equation $\phi(n)=2p$ where $p$ is prime and $2p+1$ is composite has no solutions. Using formula for $\phi$ it's quite easy proving $n$ cannot have more than two prime factors in its factorization (and we can do even better), so we must split this problem into some cases; however I'm not that familiar with number theory, so I'm not even sure which cases I must include, can you help me?  

Recall that $\phi(p^e)=p^{e-1}(p-1)$ for a prime $p$ and that $\phi$ is multiplicative (but not totally).
If $n=2p+1$ were prime, then $\phi(n)=2p$ but the question states that $2p+1$ is composite so this is not possible.
Now let $l$ be a prime such that $l\mid n$.
If $l>3$, then $l-1$ would be a factor of $2p$ which is again impossible (why?).
If $l=2$, then we can have at most $l^3\mid n$ (depending on whether $p=2$ or not).
If $l=3$, then we can have at most $l^2\mid n$ (depending on whether $p=3$ or not).
Putting this all together, we only need to check values of $n$ in the set $\{2,3,4,6,8,9,12,18,24,36,72\}$ (we could also reduce this list further but its not too bad so I won't by splitting into whether $p=2$ or $p=3$).
Calculating $\phi(n)$ for each of these, we find no solutions of the desired form.
