Integer Points of First Region as Element Set

This section describes how to use all integer points from the first region on a reduced elliptic curve as the element set to try to construct an Abelian group.

Once we know how find all integer points on an elliptic curve reduced by modular arithmetic of a prime number. We need to consider a way to define an Abelian group out those integer points.

Let's consider all integer points from the first region on a reduced elliptic curve as the element set:

```Element Set in a Single Region:
All P = (x,y), such that:
y2 = x3 + ax + b (mod p)
where:
a and b are integers
p is a prime number
4a3 + 27b2 != 0
x and y are integers in {0, 1, 2, ..., p-1}
```

Can we still use the rule of chord operation on this set as the group operation?

Let's take the same reduced elliptic curve from the previous tutorial of (a,b) = (1,4) and p = 23 as an example:

```Element Set in a Single Region:
All P = (x,y), such that:
y2 = x3 + x + 4 (mod 23)
where:
x and y are integers in {0, 1, 2, ..., 22}
```

If we want perform the point doubling operation of P = (0,2), obviously we can not do it geometrically using the rule of chord. But we do it using our algebraic equations developed earlier in the book:

```Calculation of 2P = P +P:
2P = P + P = R = (xR, yR)
P = (xP, yP) = (0, 2)

xR = m2 - 2xP                  (4)
yR = m(xP - xR) - yP           (5)

3(xP)2 + a
m = ---------                 (6)
2(yP)

m = (3*0*0 + 1)/(2*2) = 1/4
xR = (1/4)*(1/4) - 2*0 = 1/16
yR = (1/4)*(0 - 1/16) - 2 = -1/64 - 2 = -129/64

Result:
2P = R = (xR, yR) = (1/16, -129/64)
```

As you can see, the resulting point R is not an integer point and it is not in the first region! In other words, (0,2) + (0,2) using the original rule of chord operation as the addition operation does not satisfy the "Closure" condition of Abelian group.

See the next tutorial on how to reduce the point addition operation.

Last update: 2019.