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PUMaC 2015 · 加试 · 第 4 题

PUMaC 2015 — Power Round — Problem 4

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

Problem 4.4 (Reduced Rational Point; 5 ) . Let E : y + e xy + e y = x + e x + e x + e be an elliptic 1 3 2 4 6 ( ) a c curve with integer coefficients. Suppose that P = , on E is a rational point in reduced form (i.e. both b d the coordinates are reduced fractions). We may assume b and d are positive (since a or c can be negative). Give an equality relating b and d by writing one as a positive power of the other. So if you are morally convinced by now that E ( Q ) should be a group (as you should be), then you might also morally accept that Proposition 20 is also true for any elliptic curve, not just those written in short Weierstrass form. And it is! However, the definition of addition is slightly different in the general case. The reason we took P + Q as the reflection of P ∗ Q in the short Weierstrass case is that for elliptic curves written in short Weierstrass form, there is a natural horizontal line of symmetry at the x -axis. In the general case, we can find another natural horizontal line of symmetry; and so in the general case, while P ∗ Q is always the same, P + Q will be different. 2 3 2 For example, consider the curve y − 2015 y = x − 36 x + x . Notice that for all points ( a, b ) on the curve, by how left hand side is written, the point ( a, 2015 − b ) is also on the curve. Thus the horizontal line of 2015 symmetry in this case is . For this curve then, P ∗ Q is defined as the third point of intersection of the 2 2015 line P Q and the curve, and P + Q is defined as the reflection of P ∗ Q over the line y = . More generally 2 for all the curves we consider in this power round, this process of reflecting over the line of symmetry applies. 11 PUMaC 2015 Power Round Section 4 page 12 Figure 3: Horizontal Line of Symmetry. Definition 21. Let E be an elliptic curve. Then P + Q is the reflection of P ∗ Q over the horizontal line of symmetry of E . 2 3 Consider now the elliptic curve E : y + y = x − x and the point P = (0 , 0) on this curve. From this point on, we mostly refer to this curve and point P . It is the most important curve that we examine in order answer our question about prime density. 2 3

解析

Problem 4.3 (Addition Theory; 2, 8 ) . Let E : y = x + Ax + B be an elliptic curve with A, B ∈ Z and P = ( a, b ) , Q = ( c, d ) , and H = ( e, f ) be rational points on E . For simplicity, we assume the x -coordinates of P , Q , and H are distinct. a) Prove that P + Q is a rational point. b) Prove that the associative property holds. Namely, show that ( P + Q ) + H = P + ( Q + H ) . You may furthermore assume the x coordinates of P + Q and H are different, and that of P and Q + H are different as well. Proof. As an aside, I apologize to the grader of this problem for part b). a) There is a linear equation defining line P Q . Substituting this line into the E equation gives a cubic equation in terms of x . Note this cubic already has two rational solutions (the x -coordinates of P and Q ), and so it has a third rational solution. This gives a rational solution for x ( P ∗ Q ), the x -coordinate of P ∗ Q , and therefore rational y -coordinate as well. b) While there are some high-powered ways to do this, the expected way to prove this here is by algebra bash, which is straightforward. Here is a sketch. Let A = ( a, b ), B = ( c, d ), and C = ( e, f ). If any of A, B, C is O , then associativity is clear since the addition cancels out. Thus assume that all a, b, c, d, e, f ∈ Q . Since it’s clear the operation is abelian, it suffices to show ( A + B ) + C = ( C + B ) + A . For notations sake 2 3 let x ( P ) be the x -coordinate of any point P on our curve E : y = x + 2 x + 3, and similarly for y ( P ). Note that x (( A + B ) + C ) = f ( a, b, c, d, e, f ) will be a rational function in 6 variables. If we can show 1 that f ( a, b, c, d, e, f ) = f ( e, f, c, d, a, b ), this shows that x (( A + B ) + C ) = x (( C + B ) + A ). Similarly for 1 1 y (( A + B ) + C ) = f ( a, b, c, d, e, f ). 2 s − n For two general points ( m, n ) , ( r, s ) ∈ E ( Q ), note they form the line y − n = ( x − m ) and we r − m substitute into E to find the third point. By Vieta’s formula, since we know two solutions to E with 2 y substituted ( m and r ), we may just look at the x coefficient to find the third x -coordinate. Thus ( ) ( ) 2 s − n s − n x (( m, n ) + ( r, s )) = − m − r . Then y (( m, n ) + ( r, s )) = − · ( x (( m, n ) + ( r, s )) − m ) + n . r − m r − m 13 PUMaC 2015 Power Round Section 4 page 14 ( ) ( ) 2 3 d − b d − b d − b Thus the above gives us that x ( A + B ) = − a − c , y ( A + B ) = − (2 a + c ) + b , and then c − a c − a c − a   ( ) 2 3 ( ) d − b d − b ( ) 2 − (2 a + c ) + b − f c − a c − a d − b   x (( A + B ) + C ) = − − a − c − e,   ( ) 2 c − a d − b − a − c − e c − a and   ( ) 3 ( ) 3 3 d − b d − b ( ) d − b d − b 2 − (2 a + c ) + b − f − (2 a + c ) + − f c − a c − a c − a c − a d − b   y (( A + B )+ C ) = − (2 e + − a − c ) + f.   ( ) ( ) 2 2 c − a d − b d − b − a − c − e − a − c − e c − a c − a 7 6 6 5 2 5 5 2 4 2 4 By very careful expansion, we have ( A + B ) + C = (( a − a c + a e − 3 a c − 2 a ce − a e − 2 a b + 2 a bd − 4 4 3 4 2 4 2 4 2 4 4 3 4 2 3 2 3 2 3 3 3 2 a bf + 3 a c − a c e + 3 a ce + a d + 4 a df − a e + a f + 2 a b c − 2 a b e − 2 a bcd + 2 a bcf + 4 a bde + 3 4 3 3 3 2 2 3 2 3 3 3 3 2 3 2 2 2 2 2 2 2 2 2 3 a c + 4 a c e − 2 a c e − 4 a cd − 10 a cdf + 4 a ce − 4 a cf − 2 a d e + 3 a b c + 2 a b ce + a b e + 2 2 2 2 2 2 5 2 4 2 3 2 2 2 2 2 2 2 2 3 2 2 2 2 2 6 a bc f − 4 a bcde − 2 a bde − 3 a c − a c e − 2 a c e + 3 a c d + 6 a c df − 6 a c e + 6 a c f + 2 a cd e + 2 2 2 4 3 3 2 3 2 2 2 2 2 3 3 2 2 a d e + ab − 2 ab d +2 ab f − 4 ab c +2 ab c e − 2 ab ce − 6 ab df − 2 abc d − 10 abc f − 4 abc de +4 abcde + 3 2 6 5 4 2 3 2 3 3 3 3 2 2 2 2 2 4 2 abd + 6 abd f − ac − 2 ac e + 3 ac e + 2 ac d + 2 ac df + 4 ac e − 4 ac f + 2 ac d e − 2 acd e − ad − 3 4 4 3 3 3 2 4 2 3 2 2 2 2 2 2 4 4 3 2 ad f − b c + b e +2 b cd − 2 b cf − 4 b de + b c − 2 b c e + b c e +6 b cdf +6 b d e +2 bc d +4 bc f +4 bc de − 2 2 3 2 3 7 6 5 2 4 2 4 4 3 4 2 3 2 2 2 2 4 3 2 bc de − 2 bcd − 6 bcd f − 4 bd e + c + c e − c e − 2 c d − 2 c df − c e + c f − 2 c d e + c d e + cd +2 cd f + 4 6 5 5 4 2 4 4 2 3 2 3 3 3 3 2 3 2 3 2 2 2 d e ) / ( a − 2 a c + 2 a e − a c − 6 a ce + a e − 2 a b + 4 a bd + 4 a c + 4 a c e − 4 a ce − 2 a d + 2 a b c − 2 2 2 2 2 4 2 3 2 2 2 2 2 2 2 2 2 2 2 2 a b e − 4 a bcd + 4 a bde − a c + 4 a c e + 6 a c e + 2 a cd − 2 a d e + 2 ab c + 4 ab ce − 4 abc d − 8 abcde − 5 4 3 2 2 2 2 4 3 2 3 2 2 2 2 3 2 3 6 5 2 ac − 6 ac e − 4 ac e +2 ac d +4 acd e + b − 4 b d − 2 b c − 2 b c e +6 b d +4 bc d +4 bc de − 4 bd + c +2 c e + 4 2 3 2 2 2 4 9 9 9 8 8 8 7 2 7 2 7 2 7 c e − 2 c d − 2 c d e + d ) , ( a b − 2 a d − 2 a f +3 a cd +6 a cf − 3 a ef − 6 a bc − 3 a be +6 a c d +12 a cef + 7 2 6 3 6 2 6 2 6 3 6 2 6 2 6 3 6 2 6 3 6 3 6 a de − 3 a b + 9 a b d + 3 a b f + 2 a bc + 6 a bce − 6 a bd − 2 a be + 3 a bf − 10 a c d − 16 a c f − 6 2 6 2 6 3 6 2 6 3 6 2 6 3 6 3 5 2 5 2 5 2 12 a c ef − 21 a cde − a d − 6 a d f + 4 a de − 6 a df + a e f − a f − 6 a b cd − 12 a b cf + 6 a b ef + 5 4 5 2 2 5 2 5 5 3 5 2 5 5 4 5 4 5 3 12 a bc + 9 a bc e + 3 a bcd + 6 a bcdf + 6 a bce − 9 a bcf − 12 a bdef − 6 a c d + 12 a c f − 12 a c ef + 5 2 2 5 3 5 2 5 3 5 2 5 3 5 3 5 2 4 3 2 4 3 2 18 a c de + 9 a cd + 24 a cd f − 18 a cde + 27 a cdf − 6 a ce f + 6 a cf + 6 a d ef + 12 a b c + 6 a b e − 4 2 2 4 2 2 4 2 4 2 2 4 5 4 3 2 4 2 2 4 2 4 21 a b c d + 12 a b c f − 18 a b cef − 21 a b de − 6 a bc − 30 a bc e + 6 a bc d − 24 a bc df + 36 a bcdef + 4 2 2 4 5 4 5 4 4 4 3 2 4 2 3 4 2 2 4 2 3 4 2 2 24 a bd e + 12 a c d + 12 a c f + 30 a c ef + 15 a c de − 12 a c d − 33 a c d f + 30 a c de − 45 a c df + 4 2 3 4 2 3 4 2 4 3 2 3 5 3 4 3 3 3 3 3 2 3 3 2 3 3 15 a c e f − 15 a c f − 18 a cd ef − 9 a d e +3 a b − 12 a b d − 5 a b c − 9 a b ce +15 a b d − 6 a b df + 3 3 3 3 3 2 3 2 3 3 2 3 3 2 2 3 2 2 3 2 3 3 2 2 3 2 3 2 a b e − 3 a b f + 15 a b c d + 12 a b c f + 12 a b c ef + 39 a b cde − 3 a b d + 18 a b d f − 6 a b de + 3 2 2 3 6 3 4 2 3 3 2 3 3 3 3 3 3 3 2 3 2 3 2 2 9 a b df − 10 a bc +15 a bc e +15 a bc d +36 a bc df − 20 a bc e +30 a bc f − 24 a bc def − 51 a bcd e − 3 4 3 3 3 2 3 3 2 2 3 6 3 6 3 5 3 4 2 3 3 3 3 3 2 6 a bd − 18 a bd f +6 a bd e − 9 a bd f +2 a c d − 16 a c f − 12 a c ef − 30 a c de − 5 a c d +12 a c d f − 3 3 3 3 3 2 3 3 3 3 3 3 3 2 2 3 3 2 3 5 3 4 3 3 3 20 a c de + 30 a c df − 20 a c e f + 20 a c f + 12 a c d ef + 21 a cd e + 3 a d + 6 a d f − 2 a d e + 3 3 2 2 4 2 4 2 4 2 3 4 2 3 2 2 2 3 2 2 3 2 3 3 2 3 2 3 a d f +3 a b cd +6 a b cf − 3 a b ef − 12 a b c − 9 a b c e − 3 a b cd − 6 a b cdf − 6 a b ce +9 a b cf + 2 3 2 2 4 2 2 4 2 2 3 2 2 2 2 2 2 3 2 2 2 2 2 3 12 a b def + 6 a b c d − 33 a b c f + 12 a b c ef + 9 a b c de − 9 a b cd − 18 a b cd f + 18 a b cde − 2 2 2 2 2 2 2 7 2 5 2 2 4 2 2 4 2 4 3 2 4 2 2 3 27 a b cdf − 18 a b d ef +6 a bc +18 a bc e − 21 a bc d − 24 a bc df +30 a bc e − 45 a bc f − 24 a bc def + 2 2 2 2 2 4 2 3 2 2 3 2 2 2 2 3 2 7 2 6 2 5 2 9 a bc d e +15 a bcd +30 a bcd f − 18 a bcd e +27 a bcd f +12 a bd ef − 6 a c d − 12 a c ef +9 a c de + 2 4 3 2 4 2 2 4 3 2 4 3 2 3 2 2 2 3 2 2 5 2 4 2 3 3 12 a c d + 12 a c d f + 15 a c e f − 15 a c f + 12 a c d ef − 9 a c d e − 6 a cd − 12 a cd f + 6 a cd e − 2 3 2 2 4 5 2 5 2 4 2 4 2 4 4 2 3 5 3 3 2 9 a cd f − 3 a d ef − 6 ab c − 3 ab e + 15 ab c d − 12 ab c f + 6 ab cef + 15 ab de + 9 ab c + 21 ab c e − 3 2 2 3 2 3 2 3 3 2 2 3 3 2 2 2 5 2 5 2 4 9 ab c d + 30 ab c df + 6 ab c e − 9 ab c f − 24 ab cdef − 30 ab d e + 3 ab c d + 24 ab c f − 18 ab c ef − 2 3 2 2 2 3 2 2 2 2 2 3 2 2 2 2 2 2 3 2 8 6 2 51 ab c de − 3 ab c d − 18 ab c d f − 18 ab c de +27 ab c df +36 ab cd ef +30 ab d e +3 abc − 21 abc e − 5 2 5 5 3 5 2 4 3 2 2 2 4 2 3 2 2 3 6 abc d + 6 abc df − 18 abc e + 27 abc f + 36 abc def + 39 abc d e + 3 abc d − 6 abc d f + 18 abc d e − 2 2 2 3 4 2 8 7 6 2 5 2 5 3 5 2 5 3 27 abc d f − 24 abcd ef − 15 abd e +6 ac f +12 ac ef +6 ac de − 12 ac d f +6 ac de − 9 ac df − 6 ac e f + 5 3 4 2 3 3 2 2 4 2 3 3 2 3 2 4 5 2 7 6 6 6 ac f − 18 ac d ef − 9 ac d e + 6 ac d f − 6 ac d e + 9 ac d f + 6 acd ef + 3 ad e − b + 5 b d − b f + 5 3 5 2 5 2 5 4 3 4 3 4 2 4 2 4 3 4 2 3 6 3 4 2 3 b c + 3 b ce − 9 b d + 6 b df − 6 b c d + 6 b c f − 3 b c ef − 15 b cde + 5 b d − 15 b d f − b c − 9 b c e − 3 3 2 3 3 3 3 3 3 3 2 3 2 3 2 2 3 4 3 3 2 6 2 6 3 b c d − 18 b c df − 2 b c e + 3 b c f + 12 b c def + 30 b cd e + 5 b d + 20 b d f − 6 b c d − 6 b c f + 2 5 2 4 2 2 3 3 2 3 2 2 3 3 2 3 2 2 2 2 2 3 2 2 5 6 b c ef + 24 b c de + 15 b c d + 18 b c d f + 6 b c de − 9 b c df − 18 b c d ef − 30 b cd e − 9 b d − 2 4 9 7 2 6 2 6 3 6 2 5 4 2 2 3 4 3 3 3 2 3 15 b d f − 2 bc + 6 bc e + 9 bc d + 4 bc e − 6 bc f − 12 bc def − 21 bc d e − 12 bc d − 6 bc d f − 6 bc d e + 3 2 2 2 3 4 2 6 5 9 9 8 7 2 6 3 6 2 6 3 9 bc d f + 12 bc d ef + 15 bcd e + 5 bd + 6 bd f + c d − 2 c f − 3 c ef − 3 c de − 3 c d + 3 c d f − 2 c de + 6 2 6 3 6 3 5 2 4 3 2 3 5 3 3 3 3 3 2 2 4 5 2 7 6 9 3 c df + c e f − c f + 6 c d ef + 6 c d e + 3 c d + 2 c d e − 3 c d f − 3 c d ef − 3 cd e − d − d f ) / ( a − 8 8 7 7 2 6 2 6 6 3 6 2 6 2 6 2 6 3 5 2 5 2 3 a c + 3 a e − 12 a ce + 3 a e − 3 a b + 6 a bd + 8 a c + 12 a c e − 15 a ce − 3 a d + a e + 6 a b c − 6 a b e − 14 PUMaC 2015 Power Round Section 4 page 15 5 5 5 4 5 3 5 2 2 5 2 5 3 5 2 4 2 2 4 2 4 2 2 12 a bcd +12 a bde − 6 a c +12 a c e +27 a c e +6 a cd − 6 a ce − 6 a d e +3 a b c +18 a b ce − 3 a b e − 4 2 4 4 2 4 5 4 4 4 3 2 4 2 2 4 2 3 4 2 4 2 2 6 a bc d − 36 a bcde + 6 a bde − 6 a c − 30 a c e − 15 a c e + 3 a c d + 15 a c e + 18 a cd e − 3 a d e + 3 4 3 3 3 2 3 3 2 2 3 2 2 3 2 2 3 3 3 2 3 2 3 a b − 12 a b d − 12 a b c − 12 a b c e + 12 a b ce + 18 a b d + 24 a bc d + 24 a bc de − 24 a bcde − 3 3 3 6 3 5 3 4 2 3 3 2 3 3 3 3 2 2 3 2 2 3 4 2 4 12 a bd + 8 a c + 12 a c e − 15 a c e − 12 a c d − 20 a c e − 12 a c d e + 12 a cd e + 3 a d − 3 a b c + 2 4 2 3 2 3 2 2 4 2 2 3 2 2 2 2 2 2 2 2 2 2 2 4 3 a b e + 12 a b cd − 12 a b de + 3 a b c − 12 a b c e − 18 a b c e − 18 a b cd + 18 a b d e − 6 a bc d + 2 3 2 2 2 2 3 2 3 2 6 2 5 2 2 4 2 2 4 3 2 3 2 24 a bc de + 36 a bc de + 12 a bcd − 12 a bd e + 12 a c e + 27 a c e + 3 a c d + 15 a c e − 12 a c d e − 2 2 2 2 2 4 2 4 4 2 4 3 2 3 2 5 2 4 2 3 2 18 a c d e − 3 a cd + 3 a d e − 3 ab c − 6 ab ce + 12 ab c d + 24 ab cde + 6 ab c + 18 ab c e + 12 ab c e − 2 2 2 2 2 5 4 3 2 2 3 3 8 7 6 2 18 ab c d − 36 ab cd e − 12 abc d − 36 abc de − 24 abc de + 12 abc d + 24 abcd e − 3 ac − 12 ac e − 15 ac e + 5 2 5 3 4 2 3 2 2 2 4 4 6 5 4 3 4 2 4 2 3 3 6 ac d − 6 ac e + 18 ac d e + 12 ac d e − 3 ac d − 6 acd e − b + 6 b d + 3 b c + 3 b c e − 15 b d − 12 b c d − 3 2 3 3 2 6 2 5 2 4 2 2 3 2 2 2 2 2 4 6 5 4 2 12 b c de + 20 b d − 3 b c − 6 b c e − 3 b c e + 18 b c d + 18 b c d e − 15 b d + 6 bc d + 12 bc de + 6 bc de − 3 3 2 3 5 9 8 7 2 6 2 6 3 5 2 4 2 2 3 4 2 4 6 12 bc d − 12 bc d e + 6 bd + c + 3 c e + 3 c e − 3 c d + c e − 6 c d e − 3 c d e + 3 c d + 3 c d e − d )). 3 4 3 3 3 2 2 3 3 3 4 2 5 2 4 2 3 2 Similarly, we find that ( C + B )+ A = (( − a c +4 a c e − 6 a c e +4 a ce − a e − a c +3 a c e − 2 a c e + 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 5 a c d − 2 a c df − 2 a c e + a c f − 2 a cd e + 4 a cdef + 3 a ce − 2 a cef + a d e − 2 a de f − a e + 2 2 2 6 5 4 2 3 2 3 3 3 3 2 2 2 2 2 4 2 2 a e f + ac − 2 ac e − ac e − 2 ac d + 4 ac df + 4 ac e − 2 ac f + 2 ac d e − 4 ac def − ac e + 2 ac ef + 2 2 2 5 2 2 4 3 2 3 2 2 3 3 6 3 2 2 acd e − 4 acde f − 2 ace + 2 ace f + ad − 4 ad f − 2 ad e + 6 ad f + 4 ade f − 4 adf + ae − 2 ae f + 4 2 4 2 3 2 2 2 2 3 2 4 4 4 3 3 2 2 2 2 3 af + b c − 4 b c e +6 b c e − 4 b ce + b e − 2 bc d +4 bc f +2 bc de − 10 bc ef +6 bc de +6 bc e f +2 bcd − 2 3 2 3 3 3 2 4 2 4 3 7 6 6 bcd f − 10 bcde + 6 bcdf + 2 bce f − 2 bcf − 2 bd e + 6 bd ef + 4 bde − 6 bdef − 2 be f + 2 bef + c − c e − 5 2 4 2 4 4 3 4 2 3 2 3 3 4 3 2 2 2 2 2 5 2 2 2 4 3 c e − 2 c d + 2 c df + 3 c e + c f + 2 c d e − 2 c def + 3 c e − 4 c ef + 3 c d e − 3 c e + 3 c e f + cd − 3 2 3 3 3 6 3 2 4 4 3 2 4 4 3 7 4 2 2 cd f − 4 cd e − 2 cde f + 2 cdf − ce + 2 ce f − cf − d e + 2 d ef + d e + 2 de f − 2 def + e − 2 e f + 4 2 4 2 3 2 2 2 2 3 2 4 5 4 3 2 2 2 2 2 3 2 2 ef ) / ( a c − 4 a c e +6 a c e − 4 a ce + a e +2 ac − 6 ac e +4 ac e − 2 ac d +4 ac df +4 ac e − 2 ac f + 2 4 2 2 2 2 5 2 2 6 5 4 2 3 2 3 4 acd e − 8 acdef − 6 ace + 4 acef − 2 ad e + 4 ade f + 2 ae − 2 ae f + c − 2 c e − c e − 2 c d + 4 c df + 3 3 3 2 2 2 2 2 4 2 2 2 2 2 5 2 2 4 3 2 3 2 2 4 c e − 2 c f +2 c d e − 4 c def − c e +2 c ef +2 cd e − 4 cde f − 2 ce +2 ce f + d − 4 d f − 2 d e +6 d f + 3 3 6 3 2 4 3 6 3 5 3 4 2 3 3 3 3 2 4 3 5 3 6 4 de f − 4 df + e − 2 e f + f ) , ( a bc − 6 a bc e + 15 a bc e − 20 a bc e + 15 a bc e − 6 a bce + a be − 3 6 3 6 3 5 3 5 3 4 2 3 3 3 3 3 2 3 3 3 3 3 2 3 3 3 2 a c d +4 a c f +6 a c de − 18 a c ef +30 a c e f +2 a c d − 6 a c d f − 20 a c de +6 a c df − 20 a c e f − 3 3 3 3 2 3 3 2 2 3 2 4 3 2 2 3 2 3 3 3 2 3 2 2 3 5 2 a c f − 6 a c d e +18 a c d ef +30 a c de − 18 a c def +6 a c ef +6 a cd e − 18 a cd e f − 18 a cde + 3 2 2 3 5 3 2 3 3 3 3 3 2 3 3 6 3 3 2 3 6 3 3 3 2 7 18 a cde f + 6 a ce f − 6 a ce f − 2 a d e + 6 a d e f + 4 a de − 6 a de f − 2 a e f + 2 a e f − 3 a c d + 2 7 2 6 2 6 2 5 2 2 5 2 2 4 3 2 4 2 2 4 3 2 4 2 6 a c f + 6 a c de − 21 a c ef + 9 a c de + 18 a c e f + 6 a c d − 21 a c d f − 30 a c de + 24 a c df + 2 4 3 2 4 3 2 3 3 2 3 2 2 3 4 2 3 2 2 3 4 2 3 3 2 2 3 2 15 a c e f − 9 a c f − 9 a c d e +39 a c d ef +15 a c de − 51 a c def − 30 a c e f +21 a c ef − a c d e + 2 2 2 2 2 2 5 2 2 2 2 2 2 5 2 2 2 3 2 5 2 4 2 3 3 2 3 2 9 a c d e f +18 a c de +9 a c de f +9 a c e f − 9 a c e f − 3 a cd +15 a cd f +21 a cd e − 30 a cd f − 2 2 3 2 2 3 2 6 2 3 2 2 4 2 6 2 3 3 2 5 2 5 51 a cd e f + 30 a cd f − 21 a cde + 39 a cde f − 15 a cdf + 6 a ce f − 9 a ce f + 3 a cf + 3 a d e − 2 4 2 3 4 2 3 2 2 2 4 2 2 3 2 7 2 4 2 2 4 2 7 15 a d ef − 9 a d e + 30 a d ef + 24 a d e f − 30 a d ef + 6 a de − 21 a de f + 15 a def − 3 a e f + 2 4 3 2 5 8 7 6 2 5 2 5 5 3 5 2 4 2 6 a e f − 3 a ef − 3 abc + 12 abc e − 12 abc e + 6 abc d − 12 abc df − 12 abc e + 6 abc f − 18 abc d e + 4 4 4 4 2 3 2 2 3 2 3 5 3 2 2 2 4 2 3 36 abc def +30 abc e − 18 abc ef +12 abc d e − 24 abc de f − 12 abc e +12 abc e f − 3 abc d +12 abc d f + 2 2 3 2 2 2 2 3 2 3 2 6 2 3 2 2 4 4 3 12 abc d e − 18 abc d f − 24 abc de f +12 abc df − 12 abc e +12 abc e f − 3 abc f +6 abcd e − 24 abcd ef − 2 4 2 2 4 3 7 4 2 4 4 2 3 2 18 abcd e + 36 abcd ef + 36 abcde f − 24 abcdef + 12 abce − 18 abce f + 6 abcef − 3 abd e + 12 abd e f + 2 5 2 2 2 5 2 3 8 5 2 2 4 3 6 3 5 3 4 2 6 abd e − 18 abd e f − 12 abde f + 12 abde f − 3 abe + 6 abe f − 3 abe f − b c + 6 b c e − 15 b c e + 3 3 3 3 2 4 3 5 3 6 2 6 2 6 2 5 2 5 2 4 2 2 3 3 2 3 2 20 b c e − 15 b c e +6 b ce − b e +3 b c d − 6 b c f − 9 b c de +27 b c ef − 45 b c e f − 3 b c d +9 b c d f + 2 3 3 2 3 2 2 3 3 2 3 3 2 2 3 2 2 2 2 2 4 2 2 2 2 2 3 30 b c de − 9 b c df + 30 b c e f + 3 b c f + 9 b c d e − 27 b c d ef − 45 b c de + 27 b c def − 9 b c ef − 2 3 2 2 2 2 2 5 2 2 2 2 5 2 2 3 2 3 3 2 2 3 2 6 9 b cd e + 27 b cd e f + 27 b cde − 27 b cde f − 9 b ce f + 9 b ce f + 3 b d e − 9 b d e f − 6 b de + 2 3 2 2 6 2 3 3 9 8 6 2 6 3 6 2 5 2 5 5 4 9 b de f + 3 b e f − 3 b e f − 2 bc + 6 bc e + 3 bc d − 16 bc e − 6 bc f − 12 bc d e + 6 bc def + 12 bc e + 5 2 4 2 2 4 2 4 5 4 2 2 3 3 3 2 3 3 2 2 3 3 24 bc ef + 12 bc d e − 24 bc de f + 12 bc e − 33 bc e f − 6 bc d f + 12 bc d e + 18 bc d f + 36 bc de f − 3 3 3 6 3 3 2 3 4 2 4 2 3 2 2 4 2 2 2 2 4 18 bc df − 16 bc e + 12 bc e f + 6 bc f + 6 bc d e − 6 bc d ef − 33 bc d e − 18 bc d ef − 24 bc de f + 2 3 2 4 2 2 4 4 2 3 2 2 5 2 2 2 5 2 3 30 bc def + 12 bc e f − 12 bc ef − 12 bcd e + 30 bcd e f + 24 bcd e − 18 bcd e f + 6 bcde f − 6 bcde f + 8 5 2 2 4 6 5 4 3 4 2 3 3 3 3 2 6 2 3 2 6 bce − 12 bce f + 6 bce f − bd + 6 bd f + 6 bd e − 15 bd f − 18 bd e f + 20 bd f − 6 bd e + 18 bd e f − 2 4 3 3 5 9 6 2 6 9 9 8 7 2 7 2 6 3 6 2 15 bd f − 6 bde f + 6 bdf − 2 be + 3 be f − bf + c d − 2 c f + 3 c ef − 6 c de + 6 c e f − 3 c d + 9 c d f + 6 3 6 2 6 3 6 3 5 2 5 4 5 2 5 4 5 3 4 3 2 4 2 2 2 c de − 6 c df − 10 c e f − c f − 6 c d ef +12 c de +3 c def − 6 c e f +9 c ef +12 c d e − 21 c d e f − 4 5 4 2 2 4 5 4 2 3 3 5 3 4 3 3 3 3 3 2 3 2 3 3 2 3 6 c de + 6 c de f + 12 c e f − 12 c e f + 3 c d − 12 c d f − 5 c d e + 15 c d f + 15 c d e f − 3 c d f − 3 6 3 3 2 3 4 3 6 3 3 3 3 5 2 4 2 3 4 2 3 2 2 2 4 10 c de + 15 c de f − 6 c df + 2 c e f − 5 c e f + 3 c f + 3 c d ef − 12 c d e − 3 c d ef + 6 c d e f − 2 2 3 2 7 2 4 2 2 4 2 7 2 4 3 2 5 5 2 4 2 3 5 9 c d ef + 6 c de − 21 c de f + 15 c def − 6 c e f + 12 c e f − 6 c ef − 6 cd e + 15 cd e f + 9 cd e − 3 2 2 2 5 2 2 3 8 5 2 2 4 7 6 5 3 5 2 4 3 4 3 9 cd e f + 3 cd e f − 3 cd e f + 3 cde − 6 cde f + 3 cde f − d + 5 d f + 3 d e − 9 d f − 6 d e f + 5 d f − 3 6 3 3 2 3 4 2 6 2 3 3 2 5 9 6 2 3 4 6 9 6 3 3 5 d e − 3 d e f +5 d f − 6 d e f +15 d e f − 9 d f − 2 de +9 de f − 12 de f +5 df + e f − 3 e f +3 e f − 7 3 6 3 5 3 4 2 3 3 3 3 2 4 3 5 3 6 2 7 2 6 2 5 2 2 4 2 f ) / ( a c − 6 a c e +15 a c e − 20 a c e +15 a c e − 6 a ce + a e +3 a c − 15 a c e +27 a c e − 3 a c d + 2 4 2 4 3 2 4 2 2 3 2 2 3 2 3 4 2 3 2 2 2 2 2 2 2 2 6 a c df − 15 a c e − 3 a c f +12 a c d e − 24 a c def − 15 a c e +12 a c ef − 18 a c d e +36 a c de f + 2 2 5 2 2 2 2 2 2 3 2 3 2 6 2 3 2 2 2 4 2 4 2 7 27 a c e − 18 a c e f + 12 a cd e − 24 a cde f − 15 a ce + 12 a ce f − 3 a d e + 6 a de f + 3 a e − 15 PUMaC 2015 Power Round Section 4 page 16 2 4 2 8 7 6 2 5 2 5 5 3 5 2 4 2 4 4 4 3 a e f +3 ac − 12 ac e +12 ac e − 6 ac d +12 ac df +12 ac e − 6 ac f +18 ac d e − 36 ac def − 30 ac e + 4 2 3 2 2 3 2 3 5 3 2 2 2 4 2 3 2 2 3 2 2 2 18 ac ef − 12 ac d e + 24 ac de f + 12 ac e − 12 ac e f + 3 ac d − 12 ac d f − 12 ac d e + 18 ac d f + 2 3 2 3 2 6 2 3 2 2 4 4 3 2 4 2 2 24 ac de f − 12 ac df + 12 ac e − 12 ac e f + 3 ac f − 6 acd e + 24 acd ef + 18 acd e − 36 acd ef − 4 3 7 4 2 4 4 2 3 2 2 5 2 2 2 5 36 acde f + 24 acdef − 12 ace + 18 ace f − 6 acef + 3 ad e − 12 ad e f − 6 ad e + 18 ad e f + 12 ade f − 2 3 8 5 2 2 4 9 8 6 2 6 6 3 6 2 5 2 5 5 4 12 ade f + 3 ae − 6 ae f + 3 ae f + c − 3 c e − 3 c d + 6 c df + 8 c e − 3 c f + 6 c d e − 12 c def − 6 c e + 5 2 4 2 2 4 2 4 5 4 2 2 3 4 3 3 3 2 3 3 2 2 3 3 3 3 6 c ef +3 c d e − 6 c de f − 6 c e +3 c e f +3 c d − 12 c d f − 12 c d e +18 c d f +24 c de f − 12 c df + 3 6 3 3 2 3 4 2 4 2 3 2 2 4 2 2 2 2 4 2 3 2 4 2 2 4 8 c e − 12 c e f +3 c f − 3 c d e +12 c d ef +3 c d e − 18 c d ef − 6 c de f +12 c def +3 c e f − 3 c ef − 4 2 3 2 2 5 2 2 2 5 2 3 8 5 2 2 4 6 5 4 3 3 cd e +12 cd e f +6 cd e − 18 cd e f − 12 cde f +12 cde f − 3 ce +6 ce f − 3 ce f − d +6 d f +3 d e − 4 2 3 3 3 3 2 6 2 3 2 2 4 6 3 3 5 9 6 2 3 4 6 15 d f − 12 d e f +20 d f − 3 d e +18 d e f − 15 d f +6 de f − 12 de f +6 df + e − 3 e f +3 e f − f )). Factorization by Magma tells us that a factor of the numerator of x (( C + B ) + A ) − x (( A + B ) + C ) is 3 3 3 2 3 2 2 2 3 3 2 2 a c − a e − ac + ad + ae − af − b c + b e + c e − ce + cf − d e. (1) 2 3 2 3 2 3 However, note that by virtue of A, B, C ∈ E ( Q ), b = a + 2 a + 3, d = c + 2 c + 3, and f = e + 2 e + 3. Substitution of all three shows that expression 1 is in fact 0: 3 3 3 2 3 2 2 2 3 3 2 2 a c − a e − ac + ad + ae − af − b c + b e + c e − ce + cf − d e 3 3 3 3 3 3 3 3 = a c − a e − ac + a ( c + 2 c + 3) + ae − a ( e + 2 e + 3) − ( a + 2 a + 3) c + ( a + 2 a + 3) e 3 3 3 3

  • c e − ce + c ( e + 2 e + 3) − ( c + 2 c + 3) e 3 3 3 3 3 3 3 3 = a c − a e − ac + ac + 2 ac + 3 a + ae − ae − 2 ae − 3 a − a c − 2 ac − 3 c + a e + 2 ae + 3 e 3 3 3 3
  • c e − ce + ce + 2 ce + 3 c − c e − 2 ce − 3 e = 0 . Thus x (( A + B ) + C ) = x (( C + B ) + A ). Similarly, Magma shows that expression 1 is a factor of y (( A + B ) + C ) − y (( C + B ) + A ); thus y (( A + B ) + C ) = y (( C + B ) + A ). 2 3 Therefore ( A + B )+ C = A +( B + C ), and addition of rational points is associative for E : y = x +2 x +3. Therefore E ( Q ) is indeed a group. For reference, the following is code for Magma that was used to simplify the algebraic expressions. { Addition := function(x_0,y_0,x_1,y_1) x:=(y_1-y_0)^2/(x_1-x_0)^2-x_0-x_1; y:=(y_1-y_0)/(x_1-x_0)(x-x_0)+y_0; return x, -y; end function; R<a,b,c,d,e,f>:=FunctionField(Rationals(),6); x1,y1:=Addition(a,b,c,d); x2,y2:=Addition(e,f,x1,y1); m1,n1:=Addition(e,f,c,d); m2,n2:=Addition(m1,n1,a,b); Factorization(Numerator(m2-x2)); Factorization(Numerator(n2-y2)); mminusx:=a^3c - a^3e - ac^3 + a*(c^3+2c+3) + ae^3 - a*(e^3+2e+3) - (a^3+2a+3)*c
  • (a^3+2a+3)e + c^3e - ce^3 + c*(e^3+2e+3) - (c^3+2c+3)*e; } 16 PUMaC 2015 Power Round Section 4 page 17 2 3 2