The elliptic curve $y^2=x^3+3x+4$ has points O,(-1,0) and (0,2). Find five more points with rational coordinates.

The answer to this example gives: (0,-2) (5,-12) (5,12) (71/25,744/125) and (71/25,-744/125)

It seems to me that by changing the y coordinates sign you can automatically get another point, is this true?

Also I known how to calculate points P + Q, but in this case I don't see the mod explicitly expressed, do I have to solve for this?

Also is there a way to know how many points exist? Is it infinite or not?

(0,2)+(5,12) $λ=12-2/5-0=10/5$




  • $\begingroup$ The curve has rational coefficients, so group operations on rational points will give rational points. You have the identity $O$ and then two more points $P$ and $Q$. Go compute sums $P+P$, $P+Q$, $Q+Q$, sums of those, etc, and see what you can get. Your comment about changing the sign of $y$ works because $y$ shows up in the equation only as $y^2$. $\endgroup$ – aes Dec 9 '14 at 18:23
  • $\begingroup$ @aes I updated my question, could you explain how the example got the result for (0,2)+(5,12)? I am confused with how they got lambda $\endgroup$ – Math Major Dec 9 '14 at 18:30

They're using $\lambda$ as the slope of the line $\vec{PQ}$.

Namely, $\lambda = \frac{y_Q - y_P}{x_Q - x_P}$.

Now the idea is: take the line $y - y_P = \lambda (x - x_P)$ and find a third point (other than $P$ and $Q$) on this line intersected with the curve.

To do this, plug in $x = x_P + t$ and $y = y_P + \lambda t$:

$(y_P + \lambda t)^2 = (x_P + t)^3 + 3 (x_P + t) + 4$.

One solution is $t = 0$ corresponding to $P$ and one solution is $t = x_Q - x_P$ corresponding to $Q$. The third solution to this cubic gives you the third point, i.e. the point $S$ such that $P+Q+S = O$. Then $P+Q = R$, where $R = -S$ is the point such that $R + S + O = O$.

In this case, this is just $S$ with its $y$-coordinate swapped (since a vertical line passes through $S$ and this, with its "third point of intersection" at infinity, i.e. $O$).

It looks like you've been given formulas to find $R$, so you could also just use those. The proof those formulas work is by my comments above plus some algebraic manipulation.

  • $\begingroup$ i feel like this should be obvious but how can I solve for t? is there a cubic formula like there is for quadratics? $\endgroup$ – Math Major Dec 9 '14 at 19:28
  • $\begingroup$ Write it out and it will have $t$ and $(t - (x_Q - x_R))$ as factors, if you use $y_P^3 = x_P^2 + 3 x_P + 4$ and the same for $Q$, then you can divide and get the last solution $(x_R,y_R)$. Alternatively, all this work has been done for you it seems, as you have the formulas for $x_R$ and $y_R$ in your post above. $\endgroup$ – aes Dec 9 '14 at 21:05

Changing the sign of the y will indeed result in another point. Let (x, y) be a point on the curve, then y^2=x^3+ax+b=(-y)^2. Thus the point (x, -y) is also on the curve.

Finding 5 more points can be done by adding and doubling the given points until you have five more points.

There is no 'mod' in this case, since the field over which the curve is defined is the rationals (this means division is the simple division you're used to, e.g. 3/5=0.6).


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