Math 124, HW #7, Page 1, December 12, 2001  Student, F.

1. (a) D = −155

? reducedforms(-155)
[snip]
%13 = [[1, 1, 39], [3, -1, 13], [3, 1, 13], [5, 5, 9]]
? composition([1,1,39],[1,1,39])
%14 = [1, 1, 39]
? composition([3,-1,13],[3,-1,13])
%15 = [5, 5, 9]
? composition([3,-1,13],[5,5,9])
%16 = [3, 1, 13]
? composition([3,1,13],[3,-1,13])
%17 = [1, 1, 39]

We see that (1, 1, 39) is the identity element of this group and (3, −1, 13) has order 4, so this group must be isomorphic to C_{4} , the cyclic group of four elements.

(b) D = −231

? reducedforms(-231)
[snip]
%18 = [[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]]
\\ A group of order 12
? composition([1,1,58],[1,1,58])
%19 = [1, 1, 58]                       \\ Aha, the identity
? x=[2,-1,29]
%20 = [2, -1, 29]
? composition(x,x)
%21 = [4, 3, 15]
? composition(%21,x)
%22 = [8, 5, 8]
? composition(%22,x)
%23 = [4, -3, 15]
? composition(%23,x)
%24 = [2, 1, 29]
? composition(%24,x)
%25 = [1, 1, 58]             \\ So, [2,-1,29] has order 6
? y=[7,7,10]
%26 = [7, 7, 10]
? composition(y,y)
%27 = [1, 1, 58]             \\ So, [7,7,10] has order 2

This is enough to determine that the group is isomorphic to C_{2} × C_{6} = C_{2} × C_{2} × C_{3}.

(c) D = −660

? reducedforms(-660)
[snip]
%28 = [[1, 0, 165], [10, 10, 19], [11, 0, 15], [13, 4, 13], [2, 2, 83],
[3, 0, 55], [5, 0, 33], [6, 6, 29]]  \\ order 8
? x = [1,0,165]
%29 = [1, 0, 165]
? composition(x,x)
%30 = [1, 0, 165]
? y = [10,10,19]
%31 = [10, 10, 19]
? composition(y,y)
%32 = [1, 0, 165]
? z=[11,0,15]
%33 = [11, 0, 15]
? composition(z,z)
%34 = [1, 0, 165]
? w = [13,4,13]
%35 = [13, 4, 13]
? composition(w,w)
%36 = [1, 0, 165]
? v = [2,2,83]
%37 = [2, 2, 83]
? composition(v,v)
%38 = [1, 0, 165]
? xx = [3,0,55]
%39 = [3, 0, 55]
? composition(xx,xx)
%40 = [1, 0, 165]
? yy = [5,0,33]
%41 = [5, 0, 33]
? composition(yy,yy)
%42 = [1, 0, 165]
? zz = [6,6,29]
%43 = [6, 6, 29]
? composition(zz,zz)
%44 = [1, 0, 165]

All the elements have order 2, so the group is isomorphic to C_{2} × C_{2} × C_{2} .

(d) D = −12104 At this point, I used the function “orderform” from the virtual office hours and wrote my own function to determine the orders of elements.

{determineorders(v) =
        local(len,id,tmporder, orders);
        len = length(v);
        orders = [];
        \\ Let's find the identity:

        for(X=1,len, if (v[X]==composition(v[X],v[X]), id = v[X]));
        for(X=1,len,
                tmporder=orderform(v[X],id);
                orders = setunion(orders,[tmporder]);
                print(v[X], "\t has order ", tmporder);
        );
        \\ Cumulative keeping track...
        print(orders)
}

So, we used the reduced forms from the e-mail to find:

? v
%6 = [[1, 0, 3026], [10, -4, 303], [10, 4, 303], [13, -8, 234],
[13, 8, 234], [15, -14, 205], [15, -4, 202], [15, 14, 205],
[15, 4, 202], [17, 0, 178], [18, -8, 169], [18, 8, 169],
[2, 0, 1513], [25, -14, 123], [25, 14, 123], [26, -8, 117],
[26, 8, 117], [27, -10, 113], [27, 10, 113], [3, -2, 1009],
[3, 2, 1009], [30, -16, 103], [30, -4, 101], [30, 16, 103],
[30, 4, 101], [34, 0, 89], [39, -34, 85], [39, -8, 78],
[39, 34, 85], [39, 8, 78], [41, -14, 75], [41, 14, 75],
[45, -26, 71], [45, -44, 78], [45, 26, 71], [45, 44, 78],
[5, -4, 606], [5, 4, 606], [50, -36, 67], [50, 36, 67],
[51, -34, 65], [51, 34, 65], [54, -44, 65], [54, 44, 65],
[6, -4, 505], [6, 4, 505], [9, -8, 338], [9, 8, 338]]
? determineorders(v)
[1, 0, 3026]     has order 1
[10, -4, 303]    has order 12
[10, 4, 303]     has order 12
[13, -8, 234]    has order 12
[13, 8, 234]     has order 12
[15, -14, 205]   has order 4
[15, -4, 202]    has order 12
[15, 14, 205]    has order 4
[15, 4, 202]     has order 12
[17, 0, 178]     has order 2
[18, -8, 169]    has order 6
[18, 8, 169]     has order 6
[snip]
[9, -8, 338]     has order 6
[9, 8, 338]      has order 6
["1", "12", "2", "3", "4", "6"]

So we know that the group is either C_{4} × C_{12} or C_{2} × C_{2} × C_{12}. If we look at the output, there are only three elements with order 2. Therefore, we are dealing with C_{4} × C_{1}2 . (Let v generate C_{12}, and x generate C_{4}. Then (x^{2}, v^{6}) and (x^{2}, 1) and (1, v^{6}) are the only elements that have order 2. On the other hand, if x, y, and z are the order 2 elements of the Klein four group, then (x, 1), (y, 1), (z, 1), (1, v^{6}), (x, v^{6}),(y, v^{6}),(z, v^{6}) all have order three.)

(e) D = −10015 We find that there are 54 = 18 · 3 elements in this group. Furthermore, our order program has found that there exist elements of order 18. We must conclude, then, that our group is C_{18} × C_{3} .

2. See attached printouts.

3. y^{2} − x^{3} = −2 has the rational solution (3, 5). Implicitly differentiating, we have 2yy′ − 3x^{2} = 0 ⟹ y′ = \frac{3x^{2}}{2y}. So, y′(3, 5) = \frac{27}{10}. The tangent line is of the form y = \frac{27}{10}x + b and goes through the point (3, 5). We solve 5 = \frac{27}{10} · 3 + b to find b = \frac{−31}{10}. We plug y = \frac{27}{10}x − \frac{31}{10} into y^{2} − x^{3} = −2 to get: \frac{27^{2}}{10^{2}} x^{2} − \frac{2·31·27}{10^{2}}x + \frac{31^{2}}{10^{2}} = x^{3} − 2. Then, we solve this equation to find that x = \frac{129}{100}, which then implies that y = ±3831000. So we’ve found a second rational solution, namely, (\frac{129}{100}, 383, 1000). [I’ve attached a Mathematica drawing of the curve and tangent line.]