This post is a generalisation of a post I read few days ago on this site. The post asks us to find a natural number $n$ such that $\phi(n)=14$. Interestingly there are no numbers satisfying this equation. I have tried to generalize this as follows: Let $m$ be a square free even integer and $m+1$ not a prime of the form $3mod(4)$ then prove that there are no integers $n$ satisfying $$\phi(n)=m$$. I have thought a lot about this but I have no idea where to begin.
Although I do not know whether or not this is true, but I could not find a counterexample to this so I presume the statement is true.Could you please help me in proving this or bring to my notice an obvious counterexample that might have escaped me?
Thank you.
There are examples, such as $\varphi(11^2)=110=2\times 5\times 11$. (note that $111=3\times 37$ is not a prime).
Let's analyze the possibilities.
Suppose we had a counterexample $\varphi(n)=m$ where $m$ is of the form you desire.
If $n$ is divisible by any prime $p\equiv 1 \pmod 4$ then $4\,|\,p-1\implies 4\,|\,\varphi(n)$ which would tell us that $\varphi(n)$ is not square free.
Thus if $n$ were a solution to your equation we would know that $n$ could only be divisible by $2$ and odd primes $\equiv 3 \pmod 4$.
Now, if $n$ were divisible by two distinct odd primes we'd again have $4\,|\,\varphi(n)$.
Of course, if $p^3\,|\,n$ for some $n$ we'd have $p^2\,|\,\varphi(n)$
It follows from the above that $n$ must have one of the forms $$p,2p,p^2,2p^2$$ for some prime $p\equiv 3\pmod 4$.
We can eliminate the cases $p,2p$: we'd have $$\varphi(p)=\varphi(2p)=p-1$$ But that's not an acceptable $m$ by your criterion (since $m+1$ would clearly be a prime $\equiv 3 \pmod 4$).
That leaves the cases $p^2,2p^2$. There are examples along those lines, such as $m=11^2$ or $2\times 11^2$. similarly $n=19^2$ or $2\times 19^2$ gives $\varphi(n)=342$ and $343$ is not a prime.