# Are there three ordinary elliptic curves $E$, $E_1$, $E_2$ such that $E^2 \cong E_1 \!\times\! E_2$?

Consider the elliptic curve $$E\!: y^2 = x^3 + 1$$ of $$j$$-invariant $$0$$ over an algebraically closed field $$k$$ of characteristics $$p$$. Let me remind that $$E$$ is ordinary (i.e., non-supersingular) iff $$p \equiv 1$$ ($$\mathrm{mod}$$ $$3$$).

Also consider the direct square $$E^2$$. It is known that for $$p \equiv 2$$ $$(\mathrm{mod}$$ $$3$$) it is the unique superspecial abelian surface, i.e., $$E^2 \cong E_1 \!\times\! E_2$$ for any two supersingular elliptic curves $$E_1$$, $$E_2$$ over $$k$$.

What is known for $$p \equiv 1$$ $$(\mathrm{mod}$$ $$3$$)? More precisely, are there two ordinary elliptic curves $$E_1$$, $$E_2$$ over $$k$$ (at least one of them with $$j \neq 0$$) such that $$E^2 \cong E_1 \!\times\! E_2$$?