Question
Does every aperiodic 2-adic number whose corresponding real is algebraic, have a representation as a periodic p-adic for some other p?
More concrete
I'm aware that the members of two p-adic fields are incomparable unless both fields are subfields of some larger field, but I believe this circumvents that objection:
Truncate a $1:x$ rectangle shape by cutting a $1\times1$ square off its shortest end to form a new rectangle shape, and rescale up to $1:x$
Every time you truncate from the same end as you started, record a $1$, Every time you truncate from the opposite end, write a $0$.
This forms a bijection $f:\Bbb R^{\geq0}\to\Bbb Z_2$ between the real numbers greater than or equal to zero, and the 2-adic numbers. The bijection conjugates inverses in the reals to $-x$ in the 2-adics.
Since writing the 2-adic number is topologically conjugate to writing a continued fraction, it is a corollary of theorems of Lagrange and Euler that $f$ bijects the periodic 2-adic numbers with the set of all quadratic roots.
Now if periodicity fails for $p=2$, switch to the map truncating 3-hyperrectangles and 3-adics and so on.
My conjecture is that every algebraic number which is aperiodic in $\Bbb Z_2$ is periodic in some other p-adic field. Is there a proof, counterexample or reference?