tailieunhanh - Analytic Number Theory A Tribute to Gauss and Dirichlet Episode 1 Part 4

Tham khảo tài liệu 'analytic number theory a tribute to gauss and dirichlet episode 1 part 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 52 . BROWNING Y13 min But now 18 implies that Y14 B1 2 Yi YfO Y Ys 2 and 20 and 21 together imply that Yq3 c Y33Y34 Y04. We therefore deduce that X Y o1 2 X Y2 4Ý3 4Y1 2V1 2V1 4Y1 4 7 2 N B 7 2 Y03 Y04 23 24 y33 y34 Y1 Yi3 Yi4 Y03Y04Y33 20 holds Y1Y23 Y24 Y34 1 2 1 2 1 2 B 1 y y Y23 y24 Y33 y34. Finally it follows from 17 and 21 that Y33 c B1 2 Y93 2Y94 2F34 whence J X 33 X 23 24 3 N B 1 B logB 5 Y1 Yí3 Yí4 Y04 Y13 Y14 Y23 Y34 20 holds which is satisfactory for the theorem. Next we suppose that 22 holds so that 23 also holds. In this case it follows from 19 together with the inequality Y1Y13Y14 Y03Y04 that - 2 .r .V. .V1 2 Ì Y 4y3 4 Y 2 Y 4 yq4 Y14Y24 3 .- Y03y04 I 03 Yq4 Y2M Y3-1 Y1 1Y1 2 Y1Y14 I v4 24r4 24r4 4 3 23 33 1 14 Y33 On combining this with the inequality Y14 c B1 2 Y 1 2Yn4 2Y24Y34 2 that follows -L 1 04 2 34 from 18 we may therefore deduce from 25 that N Y1Y13Y14Y23Y24Y33Y34 Y1 Yí3 Yí4 Y1 Yí3 Yí4 22 holds 22 holds yỴY YỴy Y 2y3 2y3 4 Y 4 Y1 Y03 Y04 y14y23 24 y33 y34 Y33 Y34 B1 2 YA 4Yl 4Yl 2Yl 2Y3 4Y3 4 B 7 Yo3 Yo4 y23 y24 y33 y34 Y1 Y03 Y04 Y23 Y24 Y33 Y34 Now it follows from 23 that Y33 c Y03Y04 Y34. We may therefore combine this with the first inequality in 17 to conclude that N B1 2 Y03Y04Y23 2Y2Y2 B logB 5 Y1 Yi3 Yi4 Y1 Y03 Y04 22 holds Y23 Y24 Y34 which is also satisfactory for the theorem. Finally we suppose that 24 holds. On combining 19 with the fact that Y33Y34 Y03Y04 we obtain 1 Y 04Y124Í 2Y34 1 04 14 24 Y33 m N Yqyy 4Ỉr I Y 2TY1 2 TY1 2tY 1 2 Summing 25 over Y33 first with min Yo3Yo4 Y33Y34 Y03 Yo4 Y33 Y34 we therefore obtain X X Y Y1 2 1 2 3 2 1 2 3 2 1 2 4 N 4 Y1 Yo3 Y04Y13 y14 y23 y24 y34 Y1 Yi3 Yi4 Y1 Y03 Y04 Y13 24 holds Y14 Y23 Y24 Y34 AN OVERVIEW OF MANIN S CONJECTURE FOR DEL PEZZO SURFACES 53 But then we may sum over 03 Y13 satisfying the inequalities in 17 and then Y1 satisfying the second inequality in 18 in order to conclude that ss r1 4 2 1 2 3 2 1 4V5 4v1 2 s N B s Y1Y04