This is a problem from http://www.williamstein.org/ent/ent.pdf
Problem 6.2. One rational solution to the equation $y^2=x^3-2$ is $(3,5)$. Find a rational solution with $x$ not equal to $3$ by drawing the tangent line to $(3,5)$ and computing the second point of intersection.
My attempt: So I have drawn the graph, and have points $P_1=(3,5)$ and $P_2=(3,-5)$. I used (with uncertainty) the Elliptic Curve group law such that $\lambda=(3x_1^2+a)/(2y_1)$, and got $27/10$ for my value.
I then used the equation $Q=(\lambda^2-x_1-x_2, -\lambda x_3-v), v = y_1 - \lambda x_1, x_3 = \lambda^2 - x_1 - x_2$.
From this, I got the following: $Q=(1.29,-5)$.
The answer in the book says that $(129/100, 383/1000)$ is the correct answer.
Could anybody help me figure this out with a solution?
Did I coincidentally get $1.29$ for my $x$-coordinate?
There was a little mistake in the calculations. Point doubling requires $P_1=P_2=(3,5)$. Next find $\lambda=(3x_1^2+a)/(2y_1)=27/10.\;$ Now $\;x_3=\lambda^2-x_1-x_2=729/100-3-3=129/100\;$ and $\;y_3=\lambda(x_1-x_3)-y_1=27/10(3-129/100)-5=-383/1000.\;$ There is a sign difference because $P_3$ is the sum of $P_1$ and $P_2$. The second point of intersection is $(129/100,383/1000).$
You can easily check the point doubling using PARI/GP with the following code: $$\texttt{E=ellinit([0,0,0,0,-2]); print(ellmul(E,[3,5],2))}$$ with the resulting output $$\texttt{[129/100,-383/1000]}$$