# All the affine points on the curve

I have calculated the affine points on the curve $x^2 + y^2 = 1 − 3x^2y^2$ over the field ${\mathbb{Z}}_{11}.$

Using $y^2 = \frac{1-x^2}{1+3x^2}$ I got the following points:

$(0,1),(0,10),(1,0),(2,2),(2,9),(9,2),(9,9),(10,0)$

Total comes to be 8 points. But what I know is that total points should come 11-1 = 10.

So I am not getting other 2 points.

• How do you conclude that the number of points must be 11-1 ? – user27950 Oct 25 '15 at 12:18
• Are you sure about the $x^2y^2$ part? In any case, this is probably off-topic. – fkraiem Oct 25 '15 at 13:37
• I read it some where that the order normally is $p-1$ but I am not sure. Also here this curve is over $\mathbb{F}_{11}$, so I tried using 11-1 = 10 – TechJ Oct 25 '15 at 14:11
• I think you confuse that with the order of $GF(p)^*$, which is $p-1$. The order of an elliptic curve cannot be calculated so easily. The curve of your question has exactely the 8 points you stated. – user27950 Oct 25 '15 at 14:42
• This is an Edwards curve, right? Are you looking for the points over the integers, or over some finite field? – Circonflexe Oct 30 '15 at 9:18

Well, this curve has 8 affine points and I would like to share the way to find them all.

First of all, we make a square table till 10 as we have field $${\mathbb{Z}}_{11}.$$ (so 11-1=10)

Below is the table(also calculate its mod 11 values) :

  ┌───────┬─────┬────┬────┬────┬────┬────┬────┬────┬─────┬─────┬────┬
│   #   │  0  │  1 │  2 │  3 │ 4  │ 5  │ 6  │  7 │  8  │  9  │ 10 │
└───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴
│Square │  0  │  1 │  4 │  9 │ 16 │ 25 │ 36 │ 49 │  64 │  81 │ 100│
└───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴
│Mod 11 │  0  │  1 │  4 │  9 │ 5  │ 3  │ 3  │  5 │  9  │  4  │  1 │
└───────┴─────┴────┴────┴────┴────┴────┴────┴────┴─────┴─────┴────┴


Re-arranging the equation so that we have $$y$$ on the left-hand side

$$y^2 = \frac{1-x^2}{1+3x^2}$$

Now we start putting form $$x = 0$$ until $$x = 10$$ in the above equation and get the value of $$y$$ each time. If the value of $$y$$ is one of the complete squares mod 11 i.e row 3rd of the table then the corresponding value in row 1 is the $$y$$-coordinate.

$$y$$ is calculated as mod 11. Even if the value of $$y$$ is in fractions.

For $$x=0$$ we get $$y^=1$$ which exists in row 3rd i.e it is a complete square. Now value 1 in 3rd row corresponds to value $$1$$ and $$10$$ in row 1. So we get 2 points here i.e. $$(0,1)$$ and $$(0,10)$$

For $$x=1$$ we get $$y^2=0$$ this also exists in 3rd row. So corresponding values in 1st row are 0. So here we have only 1 point $$(1,0)$$

For $$x=2$$ we get $$y^2=4$$ this also exists in the 3rd row. So corresponding values in 1st row are 2 and 9. So here we have 2 points i.e. $$(2,2)$$ and $$(2,9)$$

For $$x=3$$ we get $$y^2=6$$ which does not exist in the 3rd row of the table. So no corresponding value of $$y$$ will be there, hence no point will exist at $$x=3$$

For $$x=4$$ we get $$y^2=8$$ Not in the table (not a complete square) hence no points here.

For $$x=5$$ we get $$y^2=2$$ Not in the table (not a complete square) hence no points here.

For $$x=6$$ we get $$y^2=2$$ Not in the table (not a complete square) hence no points here.

For $$x=7$$ we get $$y^2=8$$ Not in the table (not a complete square) hence no points here.

For $$x=8$$ we get $$y^2=6$$ Not in the table (not a complete square) hence no points here.

For $$x=9$$ we get $$y^2=4$$ Exists in table (hence it is a complete square) So corresponding points are: $$(9,2)$$ and $$(9,9)$$

For $$x=10$$ we get $$y^2=0$$ Exists in table (hence it is a complete square) So corresponding point is: $$(10,0)$$

So all the affine points are $$(0,1)$$ $$(0,10)$$ $$(1,0)$$ $$(2,2)(2,9)$$ $$(9,2)$$ $$(9,9)$$ $$(10,0)$$

Note that we can calculate solutions for $$x = 0, 1, ... , \frac{11-1}{2}$$ and use symmetry in $$x$$ in $$\mod 11$$ ($$10 \equiv -1 \mod 11$$, $$9 \equiv -2 \mod 11$$, etc.) as well to obtain all points, namely $$(x,y) => (x, -y) \wedge (-x, y) \wedge (-x, -y)$$.

In the example above you get all points calculating all solutions until $$x=5$$. The solution for $$x=9 \equiv -2 \mod 11$$ is obtained using symmetry for $$x=2$$ ($$(2,2) => (2, -2) = (2, 9) \wedge (-2, 2) = (9, 2) \wedge (-2, -2) = (9, 9)$$) and the solution for $$x = 10 \equiv -1 \mod 11$$ is obtained using symmetry for $$x=1$$.