Lehmer's totient problem asks if there are any composite integers $n$ with $\phi(n) \ | \ n-1$.
It is known that any such $n$ must be odd. It must also be a charmichael number.
Assume $n=4m+3$ then $\phi(n) \ | \ n-1=2(2m+1)$ . Because $n$ is a carmichael number, we have $2^3 \ | \ \phi(n)$. It follows that $2^3 \ | \ 2(2m+1)$ which is not possible therefore we must have $n=4m+1$.
If the following conjecture is true, then Lehmer's totient problem is equivalent to asking whether there are any composite integers $n$ with $\phi(n)=(n-1)/2$.
Conjecture 1. Let $n=4m+1$ where $n$ and $m$ are positive integers. Then $\phi(n) >m$.
Theorem 1. Assume Conjecture 1 holds. If $n=4m+1$ is composite and $\phi(n) \ | \ n-1$, then $\phi(n)=2m$.
Proof. Because $\phi(n) >m$ and $\phi(n) \ | \ n-1=4m$, we must have $\phi(n)= 2m$ finishing the proof.
Remark. If Conjecture 1 can be proven then we have an almost complete proof to Lehmer's totient conjecture. The possibility of $\phi(n)= 2m$ can be disproven thus providing a complete proof to Lehmer's totient problem.
There are no counterexamples to Conjecture 1 for all $m < 10^{7}$ . I need assistance in proving this conjecture or in finding at least one counterexample. For the interested readers, Conjecture 1 may be generalized to:
Conjecture 2. Let $n=am+r$ where $a$, $m$ and $r$ are positive integers with $m>a \ge 4$ and $a \ \nmid \ r$. Then $\phi(n) >m$.