 Solution

For a Short Weierstrass form $y^2=x^3+Ax+B$, use $c_4=-48A$ and $\Delta =-16(4A^3+27B^2)$. Both curves have $A=-1$, so $c_4=48=3$ in ${\mathbb{F}}_5$. For $E_1$, $B=0$ gives ${\Delta }_1=64=4$; for $E_2$, $B=1$ gives ${\Delta }_2=-368=2$. ThusThe j-invariant of an elliptic curve is preserved by every isomorphism over the algebraic closure. Since these values differ, the two curves are not geometrically isomorphic, even though the preceding part gives a degree-two elliptic isogeny.