Pairs of primes and Cardinalities of sets

For a pair of primes $\displaystyle (p_0,p_1)$, let Condition $\displaystyle A$ be the following:

For all integers $\displaystyle n>0$, $\displaystyle \mbox{card}\left( \left\{ p_{1-j}^i \in \mathbb{Z} / p_j^n \mathbb{Z} \mid i \in \mathbb{N} \right\} \right) = p_j^n-p_j^{n-1}$ for both $\displaystyle j=0$ and $\displaystyle j=1$.

Let $\displaystyle M$ be the set of all pairs of primes that satisfy Condition $\displaystyle A$. I can show that $\displaystyle (2,3) \in M$, so $\displaystyle M$ is nonempty. I think it might be true that $\displaystyle (p_0,p_1) \in M$ if $\displaystyle p_0<p_1$ and there does not exist a prime $\displaystyle p_2$ such that $\displaystyle p_0<p_2<p_1$.

Let $\displaystyle K = \left\{(p_0,p_1) \mid p_0,p_1 \mbox{ prime}, p_0<p_1, \nexists p_2 \mbox{ prime s.t. }p_0<p_2<p_1\right\}$. Then I think it may be true that $\displaystyle M = \left\{(p_0,p_1) \mid (p_0,p_1) \in K \mbox{ or } (p_1,p_0) \in K\right\}$.

Is there an obvious counterexample? (So far, I haven't found any, but I admit I haven't looked terribly hard yet).

Re: Pairs of primes and Cardinalities of sets

Ok, I figured this out. I am essentially looking for pairs of primes $\displaystyle p,q$ for which $\displaystyle p$ is a primitive root modulo $\displaystyle q^n$ for all $\displaystyle n>0$ and $\displaystyle q$ is a primitive root modulo $\displaystyle p^n$ for all $\displaystyle n>0$. I realized that $\displaystyle (2,3)\notin M$, I have no clue what I was thinking. And it is possible that $\displaystyle M$ is empty. My bad. Thanks anyway.