Let $ E: y^{2}=x^{3}+x+1 \pmod{193)} $ be an elliptic curve and $ P=(x,y) $ be a point on the curve of order $3$

Question

Let $ E: y^{2}=x^{3}+x+1 \pmod{193} $ be an elliptic curve and $ P=(x,y) $ be a point on the curve of order $3$. How do I find $x$ and $y$ if $P=(x,y)$ is of order $3$.

I know how to find the set $ E_{193}(1,1)$, but i have no idea how to find $x$ and $y$ , i.e; $(x,y)$ if the order of $(x,y)$ is $3$. Please help me.

Answer 1

This gets really messy by hand, but some code on SageMath I wrote indicates that the point $(108,66)$ and its inverse $(108, 127)$ both have order 3.

Here is the code:

E = EllipticCurve(GF(193),[0,0,0,1,1])

w = E.points()

for i in range(len(w)):
    P = w[i]
    o = P.order()
    if o == 3:
        print P, o