A. Student

Math 124 Problem Set 7

1. D=-155 There are four elements: $[[1, 1, 39], [3, -1, 13], [3, 1, 13], [5, 5, 9]].$
By the structure theorem, $\mathcal{C}_{-155}$ is isomorphic to either $C_2$x$C_2$ or $C_4$. It is easy to verify that $[1,1,39]$ is the identity. From this we find that $[3,-1,13]$ has order 4, so it must be that $\mathcal{C}_{-155}\backsimeq C_4$.
D=-231 There are twelve elements: $[1, 1, 58], [2, -1, 29], [2, 1, 29], [3, 3, 20], [4, -3, 15], [4, 3, 15],$
$[5, -3, 12], [5, 3, 12], [6, -3, 10], [6, 3, 10], [7, 7, 10], [8, 5, 8].$ Therefore $\mathcal{C}_{-231}\simeq C_{12}$ or $C_2$x$C_6$. The identity is $[1,1,58]$. Both $[2,-1,29]$ and $[2,1,29]$ have order 6, which is impossible in $C_{12}$, so $\mathcal{C}_{-231}\simeq C_2$x$C_6$.
D=-660 There are eight elements: $[1, 0, 165], [10, 10, 19], [11, 0, 15], [13, 4, 13], [2, 2, 83], [3, 0, 55], [5, 0, 33],$
$[6, 6, 29].$ The first element is the identity, and all others have order 2. Therefore $\mathcal{C}_{-660}\simeq C_2$x$C_2$x$C_2$.
D=-12104 There are forty-eight elements: (listed in an email from Professor Stein). By the structure theorem, $\mathcal{C}_D\simeq C_{48}$, $C_4$x$C_{12}$, or $C_2$x$C_{24}$. The identity element is $[1,0,3026]$, and using it we find two elements of order four: $[45,-26,71]$ and $[50,-36,67]$, eliminating everything but $C_4$x$C_{12}$.
D=-10015 There are fifty-four elements (listed in an email from Professor Stein). Therefore $\mathcal{C}_D\simeq C_3$x$C_{18}$ or $C_{54}$. The identity is $[1,1,2504]$; from this we find two elements with order 9: $[10,-5,251]$ and $[10,5,251]$. Therefore the group cannot be $C_{54}$, so $\mathcal{C}_D\simeq C_3$x$C_{18}$.
2. The three graphs are on the next page, plotted in MAPLE.
3. Differentiating implicitly, the slope of the tangent at $(x,y)$ is $\frac{3x^2}{2y}$. At $(3,5)$, the slope is $\frac{27}{10}$, and the tangent line has equation $y=\frac{27x-31}{10}$. Substituting into the relation $y^2-x^3=-2$, we have $(\frac{27x-31}{10})^2-x^3=-2$, which simplifies to the polynomial

$$100x^3-729x^2+1674x-1161=0.$$

This polynomial has a double root at $x=3$, so it factors into $(x-3)^2(100x-129)$, giving a rational root with $x=1.29$. Therefore $(1.29, .383)$ is a rational solution to the original equation.