One root defines a copy of Z2 in G. Three roots generate Z22 in G. Draw a line between any two of the three solutions, along the x axis, and find the third. This is what we expect from the nonzero elements of Z22.
This reasoning is valid for all elliptic curves, and is not specific to the rationals.
u2t3 = v2 × (s3 + ast2 + bt3)
Since u and v are coprime, v2 divides into t3. Moving to the right hand side, t does not divide the second factor, hence t3 divides v2. Put this together and v2 = t3. There is a common integer w such that v = w3 and t = w2. Rewrite our elliptic curve as follows.
u2 = s3 + asw4 + bw6
Remember that u and w are coprime.
u2 = s3 + bw6
Let's look at an example; set b = 22. Set s = 3 and w = 1 and u = 7. This gives the integers solution 3,7, which is of course a rational solution.
Set p = 3,7 and look for the order of p using the procedure described earlier. Use the tangent formula to find the slope at p, namely 27/14. This leads to 2p = -447/196,8737/2744.
The next slope, tangent to 2p, is 599427/244636. Use this to find 4p.
4p = 632287292673/59846772496, 507440701285682303/14640675036331456
At this point we can stop ( thank goodness). The x coordinate of 4p is approximately 10.5. This is sufficiently far from the origin, according to the criteria set forth in the previous section. Successive doubling, 8p, 16p, 32p, and so on, will drive the x coordinate out to infinity. The curve y2 = x3+22 defines an infinite elliptic group over the rationals, as illustrated by the point 3,7, which has infinite order.
As a corollary, there are infinitely many coprime integer solutions to the equation u2 = s3 + 22w6. Sometimes elliptic curves can be used to find and/or constrain integer solutions to high degree polynomials.
u2 = s × (s2 + aw4)
We've already analyzed u = 0, so assume u is nonzero. Now s (nonzero) divides u2. This does not mean s divides u. Let g be the gcd of s and u, and let h be the extra piece of s that goes into the "other" u. Note that h is a factor of g. In fact I'm going to replace g with g/h, so that s = gh2. Then I'm going to replace u with u/gh, so the left side becomes (ugh)2. Divide through by s and get the following.
gu2 = g2h4 + aw4
since g and w are coprime, g divides a. Select a particular value of g, divide through by g, and get this.
u2 = gh4 + (a/g)w4
When a = 1 g = ±1. Thus a sum or difference of fourth powers yields a square. This is impossible, unless one of the variables is equal to 0. We've already handled the case of u = 0, hence u h and w are nonzero, and there are no new solutions. The elliptic curve x3+x consists of the origin and ω. That's an elliptic group of Z2. The elliptic curve x3-x has two more solutions at 1,0 and -1,0, leading to the group Z22.