Lehmer’s totient problem Euler’s totient function $\varphi$ is a function defined over $\mathbb{N}$ so that $\varphi(n)=|\{m\mid m<n\wedge (m,n)=1\}|$.
Now Lehmer’s totient problem asks whether $n$ is prime iff $\varphi(n)$ divides $n-1$. 
I am curious whether the question can be expressed as a question in ring language. More specifically, whether there is a firs order formula $\psi(x)$ in the ring language so that $\mathbb{Z}\models \psi(n)$ if and only if $\varphi(n)$ divides $n-1$. 
Remark As James pointed out, the question has a positive answer. But what I really want is an algebraic answer. Probably the question was not in a proper shape. How about this: 
Whether there is a first order formula $\psi(x,y)$ in  the ring language so that there is theory $A$ extending axioms of ring theory in a proper way (which is not necessary consistent with theory of $\mathbb{Z}$) so that 
(1).  $ A\vdash\forall x\exists y\psi(x,y)\wedge (\forall x \forall y(\psi(x,y)\rightarrow( y\mbox{ divides } x-1\leftrightarrow x\mbox{ is prime})) ) $;  and
(2). For all $m$ and $n$, $\mathbb{Z}\models \psi(n,m)$ if and only if $m=\varphi(n)$.
 A: Yes. Every computable relation on $\mathbb{Z}$ can be defined with a first-order formula in the language of rings. 
The idea is to "arithmetize" computation: encode Turing machines and their states as natural numbers in such a way that the basic operations like changing the state of the head, writing a bit, etc. are all given by arithemtic relations. This is the basis of Kleene's proof of Godel's incompleteness theorem (Godel's proof used the arithmetization of syntax, rather than computation). 
A: For a more algebraic approach, we can add $\phi$ to the language of rings, and then state Lehmer's conjecture as:
$$(\text{axiom for } \phi)\ \&\ \phi(n)|(n-1) \implies Prime(n)$$
Here $Prime$ is one of a few defintions in the language of rings alone:
\begin{align}
x|y &:= \exists u \ ux=y \\
Prime(p) &:= \forall t\ t|p \implies p|t \vee t|1 \\
Power(p,q) &:= \forall t\  t|q \implies p|t \vee t|1 \\
RelPrime(q,r) &:= \forall t\ t|q\ \&\ t|r \implies t|1 \\
PosPrime(p) &:= Prime(p) \ \& \ \phi(p)=p-1
\end{align}
And the axiom for $\phi$ is the universally quantified conjunction of
\begin{align}
(p=0) \vee (p|1) &\implies \phi(p)=p\\
Prime(p) &\implies \exists u\ u|1 \ \& \ \phi(up)=up-1 \\
Power(p,q) &\implies \phi(pq)=p \phi(q) \\
RelPrime(q,r) &\implies \phi(qr)=\phi(q)\phi(r) \\
\end{align}
If $\mathbb{Z}$ satisfies the conjecture, we can ask what induction axioms are necessary to prove it.
In any case, we can ask what other rings have functions $\phi$ which satisfy the conjecture in one of these forms.
Update:  To make this more recognizable as Lehman's conjecture, we should also ensure that the axioms determine the value of $\phi$ for every element in $\mathbb{Z}$. The axioms above leave open whether $\phi(2)=2-1$ or $\phi(-2)=-2-1$, which lead to $\phi(2)=1$ or $\phi(2)=3$ respectively. The values of $\phi$ for 3 and higher primes are also independent of this axiom and this choice.
We have at least three ways to determine $\phi$ completely in $\mathbb{Z}$ if we want to:  either by 
adding the infinite set of axioms
$$\{ PosPrime(2),\ PosPrime(3),\ PosPrime(5),\ PosPrime(7),\ \ldots\}$$
or by working with ordered rings and using the axiom
$$1<p\ \&\ Prime(p) \implies PosPrime(p)$$
or by adding the single axiom 
$$PosPrime(2) \ \&\ PosPrime(3) \ \&\ PosPrime(5)
\ \&\ \Big(Prime(p+q+r) \implies$$
$$(PosPrime(p) \ \&\ PosPrime(q) \ \&\ PosPrime(r)) \implies PosPrime(p+q+r)\Big)$$
I like the single axiom, but you can decide which gives the conjecture the most ring-theoretic look.
