I have considered about Lehmer's conjecture that φ(n)|(n-1) if and only if n is a prime. I generalized the conjecture onto genernal finite commutative ring. The idea is below.


Set $A$ is a finite commutative ring, and $B$ is its unit group. Then there is a partition of $A$ by $B$ in multipication. Choose the representatives $\lbrace a\_1,a\_2,...a\_k\rbrace$, then define a subgroup $B\_i$ of $B$ as $B\_i=\lbrace b \in B \mid b\cdot a\_i=a\_i\rbrace$, for $1 \le i \le k$. Clearly, $\sum\_{i=1}^k [B:B\_i]=|A|-1$, for $\lbrace a\_i \rbrace$ is the collection of representatives. With a little more consideration, this converts to the well-known result in basic number theory that $\sum\_{d|n}\phi(d)=n$ (using Chinese residue theorem, or fact that every artin ring is a direct sum of local rings). Now, what I conjectured is, if $|B|$ can divide $|A|$, then $A$ must be a field.


I don't know much about finite commutative rings, but I can see its significance in number theory here. Moreover, I got to know some analytic methods used in the attacking of Lehmer's conjecture, such as Kevin Ford from UIUC. When faced the problem about rings, some representation theory may work. I don't know. 



A question on finite commutative rings
Suppose that $|A^\times|$ divides $|A| -1$, where $A^\times = B$ is the group of units. 


Since a finite ring has a unique factorization into local rings, we can write 
$A = \prod\_{i=1}^m A\_i$ with local rings $(A\_i,\mathfrak{m}\_i)$ and find for the unit group 


$$A^\times = \prod\_{i=1}^m \hspace{1pt} A\_i^\times.$$


[This editor makes me crazy: Without puting this formular into a single line it produces rubbish, having it in a single line, either with single-dollar or double-dollar, it works !?]


Since $(\mathfrak{m}\_i,+)$ is a subgroup of $(A\_i,+)$, we know that $|\mathfrak{m}\_i|$ divides $|A\_i|$, say $|A\_i| = k\_i |\mathfrak{m}\_i|$. Now $A\_i^\times = A\_i \setminus \mathfrak{m}\_i$ implies $|A\_i^\times| = (k\_i-1)|\mathfrak{m}\_i|$. Since $|A^\times|$ divides $|A| -1$ we conclude that $|\mathfrak{m}\_i|$ divides $-1$, what is only possible for $\mathfrak{m}\_i = 0$. Thus $A\_i$ is a field and we have shown: 



> 
> $A$ is a direct product of fields
> 
> 
> 


Let $|A\_i|=q\_i$. Then the assumption above is equivalent to 
$$ \prod\_{i=1}^m (q\_i -1) \quad \text{divides} \quad \prod\_{i=1}^m q\_i -1. \quad\quad\quad (\ast)$$


As I learned from A Haynes in the following link - and was correctly suggested by the OP - $(\ast)$ is a generalization of the Lehmer totient problem and still unsolved. 


[A question on divisibility of a product of primes](https://mathoverflow.net/questions/81816/a-question-on-divisibility-of-a-product-of-primes) 


In case $m=2$ it's easy to see that the only possibilities for $A$ are  

$$\mathbb{F}\_2 \times \mathbb{F}\_2, \quad\quad \mathbb{F}\_3 \times \mathbb{F}\_3$$
Moreover, as pointed out by François in his comment, $A= \mathbb{F}\_2^m$ satisfies the assumption for every $m$. 


