Factorization problem in cyclic cubic field Let K/$\mathbb{Q}$ be a cubic number field.  Assume that K/Q be Galois with class number 1.  
Therefore Gal(K/Q) is cyclic cubic group and $\mathcal{O}_K$ is a PID.  
Let p be a rational prime, p doesn't divide disc(K).  Then p is splits completely or inert in K.
Consider the unit group $\mathcal{O}_K^*$. Does there exists a unit $\epsilon$ in $\mathcal{O}_K^*$ such that $(\pi)$ | $(\epsilon -1)$ but $(\pi^2)$ does not divide $(\epsilon -1)$  where $\pi$ is a unramified prime over $\mathbb{Q}$?
 A: The question is slightly sloppy about quantifiers: is one fixing $K$, and then asking whether there exists such a $\pi$  (equivalently, does there exist such a $\pi$ for every $K$), or, to find a counter example, does it suffice to exhibit a single field $K$ and a single prime $\pi$ for which the statement fails? These are somewhat different questions.
Let us begin by fixing a field $K$. Since $K$ is Galois of odd order, it is totally real, and so the unit group has rank $3-1 = 2$. If $\pi$ divides the rational prime $p$, then (assuming that $p$ is unramified) the question is equivalent to asking that the image of map:
$$(1 + p \mathcal{O}_K) \cap \mathcal{O}^{\times}_K
\rightarrow (1 + p \mathcal{O}_K)/(1 + p^2 \mathcal{O}_K)$$
is trivial. (Equivalently, one is asking that the order of the ray class group of conductor $p^2$ is $p^3$ times the order of the ray class group of conductor $p$.) Actually, this is not quite correct when $p = 2$, because $-1$ is a unit of order $2$. Indeed, since if $\epsilon - 1$ is divisible by $\pi | 2$ then either $\epsilon - 1$ or $-\epsilon - 1$ is not divisible by $\pi^2$, and so $p \ge 3$.
The source of this map has rank $2$ over $\mathbf{F}_p$, and the target has rank $3$. However, this is slightly misleading; the map is Galois equivariant, and the target of the map lands inside the local units of norm one. Hence one should think of the source as being cyclic, and the target as having rank $2$. Hence, if the units are "random," one should expect that the probability that such a $p$ exists is $1/p^2$, and hence the probability that (for a fixed $K$) there are no such primes $p$ at all as being positive, since
$$\sum \frac{1}{p^2} < \infty$$
In particular, if one fixes $K$, then one certainly expects that there should only be finitely many primes with this property.
On the other hand, this will be very difficult to prove. A useful analogous problem would be to consider the case when $K = \mathbf{Q}$, and then to replace units by $S$-units for an $S$ divisible by $2$ primes. Explicitly, take $\mathbf{Z}[1/6]$-units. Then the question becomes a follows:
Does there exist a prime $p$ such that
$$2^{p-1} \equiv 1 \mod p^2, \quad 3^{p-1} \equiv 1 \mod p^2.$$
Once more, probabilistically, there should be no such $p$ (especially after one computes there are no such small $p$. Yet this is completely open. I am sure it is even unproven that the above congruence doesn't hold for all sufficiently large primes $p$. (I guess one can prove this assuming some big guns like the ABC conjecture, maybe Silverman proved this.) So the summary is:
 For any fixed cyclic cubic field $K$, one expects that only finitely many primes $p$ will have the desired property, but it will be impossible to prove this. For some fields, e.g. $K \subset \mathbf{Q}(\zeta_9)$, one would predict there are no such primes.
A second interpretation of the question is that one is also allowed to vary $K$. In this case, it is much more likely that there should exist such primes. For example, for $p = 3$ (noting $p =2$ is impossible), one might expect that (for a random $K$) the map above will be zero with probability $1/9$. Now the assumption that $K$ has class number one and $K$ is unramified at $3$ implies, by genus theory, that $K$ has prime conductor, and so lives
inside $\mathbf{Q}(\zeta_{\ell})$ for some $\ell \equiv 1 \mod 3$. So now one just has to cycle through such fields until one finds an example. (Of course, a priori, we do not even know there exist infinitely many such fields of class number one.)
The first example one finds is for $\ell = 199$. The real cubic field is given explicitly by
$$K = \mathbf{Q}(\theta)/(\theta^3 - 321\theta^2 + 119 \theta + 1)$$
One finds that $\theta^{26} \equiv 1 \mod 9$ (and so too with the other unit). This can all be confirmed in pari as follows:
(I'm not sure how to format the following correctly; please fix if you know how.)
? nf=nfinit(x^3 - 321*x^2 + 119*x + 1);
? bnf=bnfinit(nf,1);
? polgalois(nf[1])
%3 = [3, 1, 2, "A3"]
? nf[3]
%4 = 39601
? 199^2
%5 = 39601
? bnrclass(bnf,3)
%6 = [1, [], []]
? bnrclass(bnf,9)
%7 = [27, [3, 3, 3], [[3, 2, 2]~, [4, 0, 0]~, [1, 3, 0]~]]

