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 $x=y$? 

[This is an answer to the original form of the question. In the meantime the question has been clarified to refer to the multiplicative order; this seems like a much more interesting and potentially difficult question, though I'm pretty sure the answer must be yes.]
I may be missing something, but it seems the answer is a straightforward no. All non-identity elements in $\mathbb{Z} /p \mathbb{Z}$ have the same order $p$, which is different from the order $1$ of the identity element; so saying that all the orders are the same amounts to saying that $x$ and $y$ are divisible by the same primes. But different powers of the same prime, e.g. $x=2$ and $y=4$, are divisible by the same primes, and hence have the same orders.
