To show that the existential theory of $\mathbb{Z}$ in the language $\{0, 1; +, \mid , \mid_p\}$ (where $x \mid_p y \Leftrightarrow \exists r \in \mathbb{N} : y=\pm xp^r$) is undecidable we have to reduce Hilbert's tenth problem to this one, i.e., we have to reduce the existential theory of $\mathbb{Z}$ in the language $\{0, 1; +, \cdot \}$ to the existential theory of $\mathbb{Z}$ in the language $\{0, 1; +, \mid , \mid_p\}$.
Is this correct?
To do that do we have to express for example the formula $a=b \cdot c$ with the language $\{0, 1; +, \mid , \mid_p\}$ ?
We have that $$a=b \cdot c \Leftrightarrow a=\frac{(b+c)^2-(b-c)^2}{4}$$
So we have to express the formula $d=e^2$ in terms of the language $\{0, 1; +, \mid , \mid_p\}$.
Is this correct so far?
I suppose that to express $d=e^2$ we have to use $\mid_p$. I have done the following:
$$d=e^2 \tag 1$$
$$d=d_1p^l \ \ \text{ such that } d_1 \nmid p \ \ , \ \ e=e_1p^r \ \ \text{ such that } e_1 \nmid p$$
$$(1) \Rightarrow d_1p^l=e_1^2(p^r)^2=e_1^2p^{2r} \Rightarrow d_1=e_1^2p^{2r-l}$$ $$d_1 \mid e_1^2 \Rightarrow e_1^2=\lambda d_1, \lambda \in \mathbb{Z} \\ e_1 \mid d_1 \Rightarrow d_1=ke_1 , k \in \mathbb{Z}$$ $$\Rightarrow e_1^2=\lambda k e_1 \Rightarrow e_1=\lambda k$$ We would have to conclude that $d_1=e_1$ so that we can conclude that $d=ep^{l-r}$, right? But how? Or is my idea wrong?
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EDIT:
Or can we maybe just say $$d=e^2 \Leftrightarrow d \mid e =e$$ ?