Let \(X\) be the curve \(y^2=x^3-7x+10\). This curve has at least 26 points with integer coordinates. Find them (use a calculator), and verify that they are all contained in the subgroup (maybe equal to all of \(X(\QQ)\)?) generated by \(P=(1,2)\) and \(Q=(2,2)\).
Solution
Put \[ E:y^2=x^3-7x+10, \qquad P=(1,2), \qquad Q=(2,2). \]
1. The following are \(26\) integral points of \(E\):\[ \boxed{ \begin{gathered} (-3,\pm2),\quad (-2,\pm4),\quad (-1,\pm4),\quad (1,\pm2),\quad (2,\pm2),\\ (3,\pm4),\quad (5,\pm10),\quad (9,\pm26),\quad (13,\pm46),\\ (31,\pm172),\quad (41,\pm262),\quad (67,\pm548),\quad (302,\pm5248). \end{gathered} } \]
For each displayed \(x\), direct substitution in \[ x^3-7x+10 \] gives respectively \[ 2^2,4^2,4^2,2^2,2^2,4^2,10^2,26^2,46^2, 172^2,262^2,548^2,5248^2. \] Thus every displayed pair lies on \(E\), and the two signs give \(26\) distinct points.
2. The addition law needed for the subgroup check is as follows. If\[ R=(x_1,y_1), \qquad S=(x_2,y_2), \]with \(R\ne\pm S\), then\[ \lambda=\frac{y_2-y_1}{x_2-x_1}, \qquad x(R+S)=\lambda^2-x_1-x_2, \]and\[ y(R+S)=\lambda\bigl(x_1-x(R+S)\bigr)-y_1. \]For doubling,\[ \lambda=\frac{3x_1^2-7}{2y_1}. \]
These are the chord-and-tangent formulas for the short Weierstrass equation \[ y^2=x^3+Ax+B \] with \(A=-7\). They follow by intersecting the chord or tangent line with the cubic and reflecting the third intersection point across the \(x\)-axis.
3. Successive applications of the addition law give\[ \boxed{ \begin{aligned} P+Q&=(-3,-2),& 2P&=(-1,-4),\\ 3P&=(9,-26),& 2P+Q&=(3,-4),\\ 3P+Q&=(5,10),& 4P+Q&=(-2,4),\\ P-Q&=(13,46),& 2P+2Q&=(31,172),\\ 5P+2Q&=(41,-262),& 6P+Q&=(67,548),\\ -4P+Q&=(302,5248).&& \end{aligned} } \]
2. For example, the last row uses\[ Q-P=(13,-46), \qquad -3P=(9,26), \]so\[ \lambda=\frac{26-(-46)}{9-13}=-18, \]and hence\[ x=18^2-13-9=302, \qquad y=-18(13-302)+46=5248. \]This proves all the displayed identities.
4. Every point listed in step 1 belongs to the subgroup\(\langle P,Q\rangle\subseteq E(\QQ)\).
The points with first coordinates \(1\) and \(2\) are \(\pm P\) and \(\pm Q\). For the remaining first coordinates, step 3 gives one of the two points as \[ \begin{array}{c|c} x & \text{point in }\langle P,Q\rangle\\ \hline -3 & P+Q=(-3,-2)\\ -2 & 4P+Q=(-2,4)\\ -1 & 2P=(-1,-4)\\ 3 & 2P+Q=(3,-4)\\ 5 & 3P+Q=(5,10)\\ 9 & 3P=(9,-26)\\ 13 & P-Q=(13,46)\\ 31 & 2P+2Q=(31,172)\\ 41 & 5P+2Q=(41,-262)\\ 67 & 6P+Q=(67,548)\\ 302 & -4P+Q=(302,5248). \end{array} \] On a short Weierstrass equation the inverse of \((x,y)\) is \((x,-y)\). Therefore the point with the opposite sign of \(y\) is also in \(\langle P,Q\rangle\) in every row. Together with \(\pm P\) and \(\pm Q\), this accounts for all \(26\) points in step 1.
5. Q.E.D.
Step 1 finds the \(26\) integral points requested in the exercise, and step 4 verifies that every one of them lies in the subgroup generated by \(P\) and \(Q\).