Cryptography Reference
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which is not close to an integer. However,
( ω 2 / 10 + 1
2 ω 1 )= 213 . 00000 ....
This yields the point
( x, y )=( 213 , 2592)
on E . An easy check shows that this is a point of order 10. Since the order of
the torsion subgroup divides 10, we have dete rm ined that the torsion in E ( Q )
consists of the multiples of ( 213 , 2592).
Exercises
(a) Show that d 3
d 5
9.1
(mod 12) for all integers d .
(b) Show that
5
d
d 3 +7
d
d 5
0
(mod 12)
|
n
|
n
for all positive integers n .
(c) Show that
g 2 = (2 π ) 4
(1 + 240 X ) ,
12
where X = n =1 σ 3 ( n ) q n .
(d) Show that
g 3 = (2 π ) 6
(1
504 Y ) ,
216
where Y = n =1 σ 5 ( n ) q n .
(e) Show that
1728(2 π ) 12 Δ = (1 + 240 X ) 3
(1 504 Y ) 2 .
(9.41)
(f) Show that the right side of (9.41) is congruent to 144(5 X +7 Y )
mod 1728.
(g) Conclude that (2 π ) 12 Δ= n =0 d n q n ,with d n Z .
(h) Compute enough coecients to obtain that
(2 π ) 12 Δ= q (1 +
e n q n )
n =1
with e n Z .
 
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