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If $p\ge5$ then $E$ has equation $y^2=x^3+Ax+B$ with $p\mid A$ and $p\mid B$. A quadratic twist alters the discriminant, essentially $4A^3+27B^2$, by a sixth power, so for it to have good reduction $v_p(4A^3+27B^2)=6k$ where $k\in\mathbb{Z}$. Then the quadratic twist $y^2=x^3+p^{-2k}Ax+p^{-3k}B$ will work as long as $v_p(A)\ge 2k$ and $v_p(B)\ge 3k$. Otherwise any quadratic twist making the discriminant a $p$-unit will have coefficients which are non $p$-integral so no quadratic twist will have good reduction.

The cases $p=3$ or $p=2$ will be harder :-)

ADDED Even in these awkward characteristics the same argument shows that $v_p(4A^3+27B^2)$ being a multiple of $6$ is a necessary condition.

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If $p\ge5$ then $E$ has equation $y^2=x^3+Ax+B$ with $p\mid A$ and $p\mid B$. A quadratic twist alters the discriminant, essentially $4A^3+27B^2$, by a sixth power, so for it to have good reduction $v_p(4A^3+27B^2)=6k$ where $k\in\mathbb{Z}$. Then the quadratic twist $y^2=x^3+p^{-2k}Ax+p^{-3k}B$ will work as long as $v_p(A)\ge 2k$ and $v_p(B)\ge 3k$. Otherwise any quadratic twist making the discriminant a $p$-unit will have coefficients which are non $p$-integral so no quadratic twist will have good reduction.

The cases $p=3$ or $p=2$ will be harder :-)