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I think I can prove that for every $d\in\textbf{Der}_{\mathbb{Z}}(A_{\mathbb{Z}}, M_p)$, $d(A_{\mathbb{Z}}^2) = 0$. To prove it consider any $a, b\in A_{\mathbb{Z}}$ and let $n$ be a natural number, such that $p^nd(a) = p^nd(b) = 0$ (it is possible, because $M_p$ is $p$-torsion). So let $a'$, $b'$ be elements from $A_{\mathbb{Z}}$, such that $p^na' = a, p^nb'=b$ (it is possible because $A$ is $\mathbb{Q}_p$ algebra). So $d(ab) = ad(b)+d(a)b = p^na'd(b)+d(a)p^nb' = a'd(p^nb)+d(p^na)b' = 0$.
