= Solution
Let $W$ be a Brownian motion, set $X=W$, and define
$$
B_t=\int_0^t\operatorname{sign}(W_s)dW_s.
$$
This is a continuous local martingale with quadratic variation $t$, hence is Brownian by the <Lévy characterization of Brownian motion>. Since $\operatorname{sign}^2=1$,
$$
dX_t=dW_t=\operatorname{sign}(X_t)dB_t,
$$
which gives a weak solution.
Suppose a strong solution existed. It has quadratic variation $t$, so $X$ itself is Brownian. The supplied <Tanaka formula> gives $|X_t|=B_t+L_t$, and $L$ is adapted to the completed filtration of $|X|$. Hence $B$ and $|X|$ generate the same completed filtration. Strongness would make $X$, and therefore $\operatorname{sign}(X_t)$, measurable with respect to the history of $|X|$. But a Brownian excursion has an independent symmetric sign; in particular, conditionally on the reflected Brownian path, the sign at a fixed nonzero time is not measurable. This contradiction proves that \b[no strong solution exists].
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