Is an integer uniquely determined by its multiplicative order mod every prime Let $x$ and $y$ be nonzero integers and $\mathrm{ord}_p(w)$ be the multiplicative order of $w$ in $ \mathbb{Z} /p \mathbb{Z} $. If $\mathrm{ord}_p(x) = \mathrm{ord}_p(y)$ for all primes (Edit: not dividing $x$ or $y$), does this imply $x=y$? 

A related fact: integers are determined by their remainders modulo all the primes: 
Let $p_1 < p_2 < ...$ be a list of distinct primes. Then (via the Chinese remainder theorem for example) $Z_{p_1 \times ... \times p_n}$ is isomorphic to $Z_{p_1} \times .... \times Z_{p_n}$ via the map 
$$a\pmod {p_1....p_n} \rightarrow (a\pmod {p_1},....,a\pmod {p_n})$$
Thus any two distinct numbers from $1$ to $p_1....p_n$ will have distinct residues mod $p_i$ for some $i$. Thus if $a$ and $b$ are any two distinct positive integers at all, by taking $n$ large enough so that $a$ and $b$ are both less than $p_1...p_n$, there will be some $p_i$ such that $a \pmod{p_i}$ is distinct from $b \pmod{p_i}$.
