So how do you find the points on the eliptic curve $y^2=x^3+ax$ of order 4, where $4\mid a$ but $4^n$ does not divide $a$ for $n>1$. We proved that for $(x,y)=2(u,v)$, we must have $x=(u^2-a)^2/(4v^2)$. Now I wish to prove that if points $P$ with order 2 such that $P=2Q$ for some $Q$, then $a=4$.
We know that the points with order 2 are $E(2)=\{(0,0), (\pm\sqrt{a},0),\infty\}$. I tried to write $x=(u^2-a)^2/(4v^2)$ for $(x,y)\in E$, but didn't get very far. Any ideas and hints? Thanks