Math 168A: MIDTERM

William Stein

Due: Wednesday, Oct 26, 2005

The problems have equal point value, and parts of multi-part problems are of the same value. You may not talk to anybody about these problems. You are allowed to use your notes, computer software, books and web pages.

1. Make a conjecture about the set of primes p such that y^{2} = x^{3} + x has exactly p + 1 points modulo p. You do not have to prove your conjecture. You might find the following table helpful:

\begin{matrix}p & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 \\ N_{p} & 3 & 4 & 4 & 8 & 12 & 20 & 16 & 20 & 24 & 20 & 32 & 36 & 32 \end{matrix}

\begin{matrix}p & 43 & 47 & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 89 & 97 \\ N_{p} & 44 & 48 & 68 & 60 & 52 & 68 & 72 & 80 & 80 & 84 & 80 & 80 \end{matrix}

2. The point (1, 1) is a solution to the equation

3 + 2y + y^{2} = x^{2} + 2x + 3x^{3}.  (1)

Find three other solutions as follows:

(a) Multiply both sides of the equation by 9 and let X = 3x, Y = 3y to obtain an equation of the form

Y^{2} + a_{1}XY + a_{3}Y = X^{3} + a_{2}X^{2} + a_{4}X + a_{6}.  (2)

(b) Find the point P on (2) corresponding to the point (1, 1) on (1).

(c) Use the group operation on (2), e.g., via SAGE, to compute several multiples of P.

(d) Use the multiples of P to recover points on (1).

3. The right triangle with rational side lengths a = 24/5, b = 35/12, c = 337/60 has area 7. Find another right triangle with rational side lengths and area 7 as follows (note that none of the sides lengths of this second triangle are allowed to be the same as the side lengths of the above triangle):

(a) Find the point P on y^{2} = x^{3} − 49x corresponding to the triangle (24/5, 35/12, 337/60) under the bijection from class (see notes from 2005-10-17).

(b) Compute Q = P + P using the elliptic curve group law.

(c) Find the triangle corresponding to Q under the bijection from class.