\\ Included here are the commands needed for the \\ computations in the paper \\ "On a Shimura Curve that is a Counterexample to the Hasse Principle" \\ by Siksek & Skorobogatov \\ The reader is meant to run gp and copy and paste \\ the commands to verify the computation \\ The following inputs the file PariRoutines \r PariRoutines.gp \\ w will equal \sqrt{-13} w=quadgen(-13*4) g=-(3*u^2+12*u+13)*(u^2+12*u+39) \\ We will represent the curve Y: v^2=g(u) (Equation 2 in paper) \\ by a vector of coefficients [a,b,c,d,e] Y=Vec(g) P_0=[(-39+4*w)/7,(260-120*w)/49] \\ A check to see if the point P_0 is on the quartic curve Y isOnQuartic(Y,P_0) \\ The following command generates a quintuple of the form \\ Quint=[Y,P_0,ell,Phi,Psi] \\ where Y, P_0 are as above \\ ell is a Weierstrass elliptic curve, \\ Phi: Y ----> ell taking P_0 to infinity \\ Psi is the inverse map \\ Here we are thinking of Y as an equation in u,v (as in the paper) \\ and ell=[a1,a2,a3,a4,a6,....] as an equation in x,y : \\ y^2+a1*x*y+a3*y=x^3+a2*x^2+a4*x+a6 \\ Thus Phi=[x,y] where x,y are given in terms of u,v \\ and Psi=[u,v] where u,v are given in terms of x,y Quint=quarToEll2(Y,P_0) \\ The following changes of coordinates simplify the elliptic curve \\ while updating the maps Phi,Psi so that the above remain true Quint=VeryShort(Quint) Quint=ChangCoord(Quint,[2/3,0,0,0]) Quint=ChangCoord(Quint,[1,-3,0,0]) Quint=ChangCoord(Quint,[3,0,0,0]) \\ Now Quint=[Y,P_0,E,Phi,Psi] where E is the elliptic \\ curve y^2=(x-10)*(x+3)*(x+6) given in the paper \\ Phi: Y ---> E is the birational map taking P_0 to infinity \\ and Psi is the inverse map \\ Below are the points S_1,S_2,S_3 of Lemma 2.1 \\ that form a basis for E(K) S_1=[10,0] S_2=[-3,0] S_3=[-14/13,(480/169)*w] E=Quint[3] ellisoncurve(E,S_1) ellisoncurve(E,S_2) ellisoncurve(E,S_3) \\ We want now to check that S_1, S_2, S_3 are \\ independent in E(K) where $K=\Q(\sqrt{-13})$ \\ as claimed in the proof of Lemma 2 \\ We will do this using the injective homomorphism \\ E(K)/2E(K) ----> K^*/(K^*)^2 \times K^*/(K^*)^2 \times K^*/(K^*)^2 \\ [x,y] ----> [x-10,x+3,x+6] mod squares \\ with suitable patching for points of order 2 \\ (See Cassels, Lectures on Elliptic Curves, page 67) \\ We will do this calculation by hand! \\ Note S_1 ---> [ ? , 13 , 16] = [? , -1 , 1] = [-1, -1, 1] \\ S_2 ---> [-13 , ? , 3] = [1 , ? , 3] = [1 , 3 , 3] \\ S_3 ---> [-144/13 , 25/13, 64/13] = [1,-1,-1] \\ Now simply note that [-1,-1,1] , [1,3,3], [1,-1,-1] \\ are independent modulo K^*^2 \\ The command MapEllToQuar maps a point on E \\ to a point on Y \\ P_1, P_2, P_3 are the points in paper \\ mentioned just after Lemma 2 P_1=MapEllToQuar(Quint,S_1) P_2=MapEllToQuar(Quint,S_2) P_3=MapEllToQuar(Quint,S_3) \\ The following check that P_1,P_2,P_3 \\ really are on the quartic Y isOnQuartic(Y,P_1) isOnQuartic(Y,P_2) isOnQuartic(Y,P_3)