number theory 2
Tài liệu Frontiers in Number Theory, Physics, and Geometry II docx
Ngày tải lên :
12/02/2014, 16:20
... 303 4 72 479 491 555 583
T 21 121 121 226 1··· 22 5 141 328 34
(the omitted values contain only the primes 2 and 3; 3 occurs whenever d ≡ 3
mod 9 and there is also some regularity in the powers of 2 occurring). ... →+∞
L(x)=2L(1) =
π
2
3
, lim
x →−∞
L(x)=−L(1) = −
π
2
6
,
but it is continuous if we consider it modulo π
2
/2. Moreover, the new function
L(x):=L(x) (mod π
2
/2) from P
1
(R)toR/
π
2
2
Z, just ... ,r
2
). Then the deter-
minant of this matrix is a non-zero rational multiple of |d|
1 /2
ζ
F
(2) /π
2r
1
+2r
2
.
If instead of taking any r
2
linearly independent elements we choose the ξ
j
to
2
It...
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- 549
- 1
Tài liệu Frontiers in Number Theory, Physics, and Geometry I ppt
Ngày tải lên :
12/02/2014, 17:20
... −e
−l
p
n
=
p.p.o.
l
p
∞
n=1
1
2sinhnl
p
/2
e
−l
p
n(s−1 /2)
.
Choose α = s 1/2and = s
1/2then
n
1
k
2
n
+(s 1 /2)
2
1
k
2
n
+(s
1 /2)
2
=
à(D)
4
k tanh k
1
k
2
+(s − 1 /2)
2
−
1
k
2
+(s
− 1 /2)
2
dk
+
1
2s ... function
h(k)=
1
k
2
+ α
2
−
1
k
2
+ β
2
.
Its Fourier transform is
g(l)=
1
2
e
−α|l|
−
1
2
e
−β|l|
.
The Selberg trace formula gives
n
1
k
2
n
+
2
1
k
2
n
+
2
=
à(D)
2
k tanh k
1
k
2
+
2
1
k
2
+ ... compact
8 Eugene Bogomolny
k
1
=
2
a
n
1
,k
2
=
2
b
n
2
,
with n
1
,n
2
=0, ±1, 2, , and, consequently, energy eigenvalues are
E
n
1
n
2
=
2
a
n
1
2
+
2
b
n
2
2
. (3)
The first step of construction...
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Elementary Number Theory: Primes, Congruences, and Secrets pdf
Ngày tải lên :
07/03/2014, 16:20
... 106603488380168454 820 927 220 3600 128 7867 920 795857598 929
1 522 27060 823 71930 628 08643.
The next RSA challenge is RSA-640:
310741 824 0490043 721 350750035888567930037346 022 8 427 27545 720 1619
48 823 206440518081504556346 829 671 723 2867 824 3791 627 2838033415471
073108501919548 529 007337 724 822 783 525 7 423 864540146917366 024 7765
23 46609,
and ... $30,000:
740375634795617 128 28046796097 429 5731 425 9318888 923 128 908493 623 2
6389 727 65034 028 26 627 6891996419 625 1178439958943305 021 2758537011
896809 828 673317 327 31089309005 525 0511687706 329 90 723 963807867100
860969 625 37934650563796359
SAGE ... 155-digit number
10941738641570 527 421 809707 322 0403576 120 037 329 4544 920 599091
38 421 3147634998 428 893478471799 725 789 126 73 324 97 625 7 528 99781
83379707653 724 4 027 146743531593354333897.
It was factored on 22 August...
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Solved and unsolved problems in number theory daniel shanks
Ngày tải lên :
15/03/2014, 16:03
... to perfect numbers.
The first four perfect numbers are
2( 2*
-
l),
22 (23
-
11,
24 (25
-
I),
26 (27
-
1).
We raise again Conjecture
1.
Are there infinitely many perfect numbers?
We ...
Fermat Numbers,
F,
=
22 m
+
1
for
m
=
0,
1,
2,
. .
. ,
are prime to each other, since
F,+1
==
FoFl
. .
.
Fm
+
2.
Here, and throughout this book,
22 m
means
2( 2m),
not
(2& apos;) ...
zl
,
x2
,
.
.
.
,
zn
. Let
b
have at least
one,
yl
. Then each element
y.
=
y,xl-lx,
( 120 )
2 -2
2
for
i
=
1,
2, .
.
.
,
n
satisfies
7/ 12
=
11
since
2/ 12
=
11,...
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104 Number Theory Problems ppt
Ngày tải lên :
30/03/2014, 02:20
... odd, then n
2
+
n
2
−
1
2
and n
2
+
n
2
+
1
2
are integers. Clearly, we have
n
2
+
n
2
−
1
2
2
< n
4
+ n
3
+ n
2
+ n + 1.
Note that
n
2
+
n
2
+
1
2
2
= n
4
+ n
3
+
5n
2
4
+
n
2
+
1
4
= ... such that 10
m
20 05!. Since 10
m
=
2
m
5
m
,wehavem = min{e
2
(20 05!), e
5
(20 05!)}. Because 2 < 5, we have
m = e
5
(20 05!) =
20 05
5
+
20 05
25
+
20 05
125
+
20 05
625
= 500.
The ... 1.
If n is even, then n
2
+
n
2
and n
2
+
n
2
+1 are two consecutive integers. We have
n
2
+
n
2
2
= n
4
+ n
3
+
n
2
4
< n
4
+ n
3
+ n
2
+ n + 1
<
n
2
+
n
2
+ 1
2
.
Hence 11111
(n)
is...
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- 2
a course in number theory and cryptography 2 ed - neal koblitz
Ngày tải lên :
31/03/2014, 16:20
... correspond to the integers
354, 622 and 20 3, respectively
-
we obtain the integers 365, 724 and 24 .
Writing 365
=
13 -27 +14, 724
=
26 .27 +22 ,
24
=
0 .27 +24 , we put together
the plaintext ... fifth row
is
2 2 2 2 2
-
2
,
and this gives the congruence ( 42 -43 61
.
85 )2
=
(2
-3
-5
7
.
13)~ mod n, i.e., 145g2
=
90 12 mod 1 829 . We conclude that a
factor of 1 829 is g.c.d.(1459 ...
3t1,1 )2
3
10 423 87
=
7 (mod 9), i.e., 6tlYl
=
6
(mod 9), i.e.,
2tl,1
2 (mod3), so tlYl
=
1. Next, modulo 27 : (1+3+9t~ ,2) ~
=
10 423 87
=
25 (mod 27 ), i.e., 16
+
l8tlY2
-
25 (mod 27 ),...
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algebra and number theory - baker a.
Ngày tải lên :
31/03/2014, 16:21
... b = 72 we have
190 = 2 ì 72 + 46, 46 = 190 + (2) ì 72,
72 = 1 ì 46 + 26 , 26 = 72 + (1) ì 46,
46 = 1 ì 26 + 20 , 20 = 46 + (1) ì 26 ,
26 = 1 ì 20 + 6, 6 = 26 + (1) ì 20 ,
20 = 3 ì 6 + 2, 2 = 20 + ... 3 ì 2, 2 = gcd(190, 72) .
Working back we nd
2 = 20 + (3) ì 6
= 20 + (3) ì (26 + (1) ì 20 ),
= (3) ì 26 + 4 × 20 ,
= (−3) × 26 + 4 × (46 + (1) ì 26 ),
= 4 ì 46 + (7) ì 26 ,
= 4 ì 46 + (7) ì ( 72 + ... (7) ì 72 + 11 ì 46,
= (7) ì 72 + 11 ì (190 + (2) ì 72) ,
= 11 × 190 + ( 29 ) × 72.
Thus gcd(190, 72) = 2 = 11 ì 190 + (29 ) ì 72.
From this we obtain gcd(190, 72) = 2 = 11 ì 190 + 29 ì ( 72) .
It...
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algebraic groups and number theory - platonov & rapinchuk
Ngày tải lên :
31/03/2014, 16:21
... factorization g
=
klesk2 as
in (3.17). Then by properties (b) and (c) of the absolute value function
we have
~~+~~1+ ~2
Ad(s)L
c
L, whence nnf2c1+c2~
c
Ad(s)L, and finally,
nn+4c1+2c2 L
C
Ad(g) ... isomorphism (VL, xL)
21
(WL, yL). As in
52. 2.1, there is a map
PROPOSITION
2. 5.
cp
is
a bijection.
We need only prove the surjectivity of cp, for which we use
LEMMA 2. 2.
H1 (3, GLn(L)) ... sequence (2. 7) as
It follows from Lemma 2. 2 that H1(K, GL,)
=
1. But det: GLn(K)
-+
K* is surjective. Therefore, a consequence of (2. 8) is
Moreover, the special case of Lemma 2. 2 for n
=...
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algebraic number theory - iyanaga
Ngày tải lên :
31/03/2014, 16:21
...
and (9) imply
c
1
=
la0j2 sin 0 /2. Hence lim Rv(o;'oo,u[)
=
R-m
v;'
j~,l-~
(sin 8 /2) -'
=
v,
(sin 8 !2) -', (8
#
0).
(q.e.d.)
12
J.
COATE~
and
A.
WILES ...
dr'
=
21
Sr
Jr
Jm(2rrn)amn(rr')rp+2n-3 dr rh-
f
which is equal to
Now, the last two integrals exist in the region determined,
for instance, by
O<Res<& and -2n+3-&<Rep< ...
(1975), 185 -21 0.
[HI
-
,
Arithmetic and Galois module structure, for tame extensions, Crelle 28 61&apos ;28 7
(1976), 380-440.
[F4]
-
,
Galois module structure, Algebraic Number fields,...
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algebraic number theory - milne
Ngày tải lên :
31/03/2014, 16:21
... Zα
2
.There
are the following factorizations:
2( 1+X)(1 + X + X
2
) (2) = (2, 1+α) (2, 1+α + α
2
)
3 (2+ X) (2 + X + X
2
)(3) =(3, 2+ α)(3, 2+ α + α
2
)
5(1+X)(1 + 4X + X
2
)(5) =(5, 1+α)(5, 1+4α + α
2
)
7 ... + Zα
2
.There
are the following factorizations:
2( 1+X)(1 + X + X
2
) (2) = (2, 1+α) (2, 1+α + α
2
)
3 X(1 + X
2
)(3)=(3,α)(3, 1+α
2
)
5 X (2 + X
2
)(5)=(5,α)(5, 2+ α
2
)
7 (5 + X)(3 + 2X + X
2
)(7) ... ideal.
4 027 (22 15 + X)
2
(3 624 + X) (4 027 ) = (4 027 , 22 15 + α)
2
(4 027 , 3 624 + α).
Example 3.50. Let α be a root of X
3
−8X + 15. Here again, the discriminant of
the polynomial is −4 027 , and so the ring...
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analytic number theory - newman d.j.
Ngày tải lên :
31/03/2014, 16:21
... Analytic Number Theory
and
1
(1 − z)
3
d
dz
1
2( 1 − z)
2
d
dz
n + 1
2
z
n
(n + 2) (n + 1)
2
z
n
,
C(n)
(n + 2) (n + 1)
12
+
n + 1
4
+
χ
1
(n)
4
+
χ
2
(n)
3
(2)
where χ
1
(n) 1if2|n and ... I
a,b,N
,
N
n1
e(xn
k
)
≤
C
2
N
(b + j)
.
Erd
˝
os–Fuchs Theorem 35
C
1−r
2
,A
2
(r
2
) ≥ P(r
2
) +
C
1−r
2
,A
2
(r
2
) ≥−M +
C
1−r
2
, and finally
A(r
2
) ≥
−M +
C
1 − r
2
,(10)
a rate of growth ... in this same region, setting
φ(z)
1 − z
2
exp
π
2
12
1 + z
1 − z
∞
n0
q(n)z
n
,(10)
φ(z)
2
2π
exp
π
2
12
2
1 − z
exp
π
2
12
2
3(1 −|z|)
so that
φ(z) exp
1
1 −|z|
when
|1...
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analytic number theory- jia & matsumoto
Ngày tải lên :
31/03/2014, 16:21
... Theorem, Asterisque,
24 -25
(l975), 28 1 -29 3.
[6] P. M. Ross, On Chen's theorem that each large even number has
the form pl +p2 or pl +p2p3, J. London. Math. Soc. (2)
10
(1975),
500-506. ...
C,(u) inductively for
r
2
2 by
Lemma
2. 2.
Let
B
and
6
be fixed positive numbers,
k
be a natural
number, and X be suficiently large number. Suppose that z
2
x6, and
that
a
=
a/q ...
bound
Now it follows from
(4 .20 ), (4 .22 ) and the last inequality that
which gives the lemma.
1 82
ANALYTIC NUMBER THEORY
[7] R. A. Rankin, Contribution to the theory of Ramanujan's function...
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