Fun with Certificates part II - Institute for Advanced Study · Fun with Certificates part II Elliptic Curve Cryptography May 13, 2019 Network Security Institute for Advanced Study
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Fun with Certificates part IIa Deep Dive into Elliptic Curve Cryptography for all ages
Brian Epstein(he/him/his)
Institute for Advanced Study
Computer Manager, Network and Security
Information Security Officer
bepstein@ias.edu - @epepepep
https://security.ias.edu 3
Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Topics● Explain why ECC came about● ECC deep dive● Safe Curves and Trust● Certs● Demo
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
So I was browsing the Interwebs...
5
6
https://security.ias.edu 7
Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
26%
1%
64%
7%1%
no SSL
RSA 1024 bit
RSA 2048 bit
RSA 4096 bit
ECC 256 bit
.edu's taken from Majestic's top 1 million websites (3096 total)
2017 TLS Certificate Breakdown for Edu's
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
13%
0%
78%
6%
3%
0%
no SSL
RSA 1024 bit
RSA 2048 bit
RSA 4096 bit
ECC 256 bit
ECC 384 bit
.edu's taken from Majestic's top 1 million websites (4008 total)
2018 TLS Certificate Breakdown for Edu's
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Why create ECC, we have RSA?
● If RSA breaks, what then?● Faster computers force increased key size● Speed is faster with ECC (for most things)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Key Length Comparison
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Elliptic Curve Cryptography
Neal Koblitz
Victor Miller
1985
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Elliptic Curve Cryptography (ECC)
● Explain the end goal for ECC● Review a little math● Show how to get to our end goal
So, let's begin at the end...
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Secret Exchange
Where can wetalk privately??I have an
idea . . .
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
https://security.ias.edu 15
Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo (2,38)
gazelle (21,31)
stallion (17,15)
donkey (30,35)
iguana (19,18)
jackal (15,40)
orangutan (33,14)
goat (38,15)
mongoose (32,29)
rat (28,19)
deer (27,26)
cow (10,38)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29,3
8)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29
,38)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
G = dingo (2,38)
d = ?
Q = wombat (22,16)
r = ?
R = panda (26,28)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29
,38)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
G = dingo (2,38)
d = ?
Q = wombat (22,16)
r = ?
R = panda (26,28)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29
,38)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
G = dingo (2,38)
d = ?
Q = wombat (22,16)
r = ?
R = panda (26,28)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29
,38)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
G = dingo (2,38)
d = ?
Q = wombat (22,16)
r = ?
R = panda (26,28)
S = mule (21,10)
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
G = dingo (2,38)
d = ?
Q = wombat (22,16)
r = ?
R = panda (26,28)
S = mule (21,10)
d r
+ r+ d = = SWhy does this work?
d = 16 r = 25
25 + 16 = 16 + 25 = 41
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Humpf,how romantic...
I should'vepicked a better
number.
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Math● Square and Square root● Graphing● Elliptic Curves with point math● Finite Fields
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Square and Square Root
√9=3
√9=−3√9=±3
32=3⋅3=9
(−3)2=−3⋅−3=9
32=3⋅332
(−3)2=−3⋅−3(−3)2
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Graphing
0 1 2 3 40
2
4
6
8
10
12
14
16
18
1
x
y
y=x2+1
x yx2+1
0
1
2
34
5
02+1
1017
26
0•0+10+11
12+11•1+11+12
22+12•2+14+15
0 1 2 3 40
2
4
6
8
10
12
14
16
18
12
x
y
0 1 2 3 40
2
4
6
8
10
12
14
16
18
12
5
x
y
0 1 2 3 40
2
4
6
8
10
12
14
16
18
12
5
10
x
y
0 1 2 3 40
2
4
6
8
10
12
14
16
18
12
5
10
17
x
y
0 1 2 3 4 50
5
10
15
20
25
30
12
5
10
17
26
x
y
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Elliptic Curves
Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
An Elliptical Machine
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Elliptic Curvesy2= x3−x+3{(x , y )∈ℝ2∣y2= x3+ax+b ,4 a3+27 b2≠0}∪{0}
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Elliptic Curve Math● Create “point addition” ⊕
P⊕Q⊕R=0
P⊕Q=-R● Create “point multiplication” ⊙
2⊙P = P⊕P
5⊙P = P⊕P⊕P⊕P⊕P● Demo D
32
33
34
35
36
37
38
39
40
41
42
43
44
45
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
0 1 2 3 40
2
4
6
8
10
12
14
16
18
12
5
10
17
x
y
x x2+1
0
1
2
34
5
1017
26
0
2
5
Graphing
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Finite Fields● Finite
– There is an end● Field
– Football– Soccer
● Demo A
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Benefits from Finite Fields● computers are terrible at irrational
numbers● get to use whole numbers (integers)● reduce the size of the problem● Field is “closed”
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Example Finite Field● Field size is 41● x axis goes from 0 to 40● y axis goes from 0 to 40
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
Graphing an EC on a Finite Field
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2 3;
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2 3; 38
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
5
3; 38
0
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
6
3; 38
7; 34
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
... …
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0; 41
y2(mod 41)≡ x3−x+3 (mod 41)x
... …
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
... …
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
... …
https://security.ias.edu 61
Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
Point Addition ⊕● Draw a line between points P and Q● Flip over at the sides, keep your slope● When you hit the next point, flip to
opposite side of the graph
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
... …
P
Q
R
-R
P ⊕ Q = -RP ⊕ Q ⊕ R = 0
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Fun with Certificates part IIElliptic Curve Cryptography
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Network SecurityInstitute for Advanced Study
One way function
X
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
One way function● Point addition ⊕ and multiplication ⊙ are easy● Point subtraction ⊖ and division ⊘ are hard● Given R, what are P & Q?
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Fun with Certificates part IIElliptic Curve Cryptography
May 13, 2019
Network SecurityInstitute for Advanced Study
0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930313233343536373839400
5
10
15
20
25
30
35
40
x
y
2
610
1517
18
1; 4015; 26
6; 35
3; 38
7; 343; 38
5 0
y2(mod 41)≡ x3−x+3 (mod 41)x
... …R
P + Q = -R
P + Q + R = 0
-R
P
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Point Multiplication ⊙
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2 ⊙ (2,38) =(2,38) ⊕ (2,38) =(21,31)
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3 ⊙ (2,38) =(2,38) ⊕ ((2,38) ⊕ (2,38)) =(2,38) ⊕ (21,31) =(17,15)
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4 ⊙ (2,38) =(2,38) ⊕ ((2,38) ⊕ ((2,38) ⊕ (2,38))) =(2,38) ⊕ ((2,38) ⊕ (21,31)) =(2,38) ⊕ (17,15) =(30,35)
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dingo(2,38)
gazelle
(21,31)
stallio
n
(17,15)
donke
y
(30,
35)
chipmunk(2,3)
mule(21,10)alligator(17,26)
jack
al(1
5,4
0)
igua
na(1
9,18
)
ferret
(19,2
3)
boar
(30,6)
ora
ng
uta
n(3
3,1
4)
goat
(38,1
5)
mong
oose
(32,2
9)
rat
(28
,19)
deer
(27,2
6)
cow
(10
,38)
hip
po
(29
,3)
meeka
t(6
,34
)
musk
rat
(34,3
5)
cou
gar
(37
,5)
wom
bat
(22
,16)
newt(35,11)
gibbon
(25,33)
opossum
(18,35)
panda
(26,28)
llama(5,0)
sloth (35,30) pa
rake
et(2
5,8)
reindee
r(3
7,3
6)
dormouse
(22,25)
hed
geh
og
(26,1
3)
chim
panz
ee
(18,
6)
cat
(34,6
)
an
teate
r(6
,7)
oce
lot
(29
,38)
beaver
(10,3
)
pon
y(2
7,1
5)
porcu
pin
e(2
8,2
2)
squ
irrel
(32,1
2)
koala
(38
,26)
an
telo
pe
(33,2
7)
chicke
n(1
5,1
)
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And they lived happily ever after...
… until Dual_EC_DRBG
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Dual Elliptic Curve Deterministic Random Bit Generator (Dual_EC_DRBG)● Developed prior to 2004 by NSA
● Approved by NIST
● RSA Security used as default
● Bruce Schneier concluded it weak
● Edward Snowden leaks included documents revealing plot by NSA
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Is ECC compromised then?● No, but we have some trust issues.● ANSI X9.62 (1999), IEEE P1363 (2000)?● SEC 2 (2000), NIST FIPS 186-2 (2000)?● ANSI X9.63 (2001), Brainpool (2005)?● NSA Suite B (2005)?● ANSSI FRP256V1 (2011)?
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SafeCurves
● Choosing safe curves for elliptic-curve cryptography
● https://safecurves.cr.yp.to/
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Million Dollar ECC curve
● Publicly verifiable randomness produced in February 2016 by many national lotteries
● http://cryptoexperts.github.io/million-dollar-curve/
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And they lived happily ever after...
Hello? It's Eve here.Did you forget about me?
I heard everything!
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(Jimmy)nkwwm
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nkwwm
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nkwwmJimmy (Jimmy)orqql
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Jimmy orqql
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Jimmy orqqlJimmy
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Jimmy Jimmy
(got it)ldg jg
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Jimmy Jimmy
ldg jg
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Jimmy Jimmy
ldg jggot it(got it)tes fs
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Jimmy Jimmy
got ittes fs
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Jimmy Jimmy
got ittes fsgot it
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RSA Certificates● Subject (FQDN)● Issuer (CA)● Public Key
● Modulus (n) product of two prime numbers● Public Exponent (e)
● x509 extensions● Certificate Authority Signature
2008-05-29 Fun with Certificates
2008-05-29 Fun with Certificates
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ECC Certificates● Subject (FQDN)● Issuer (CA)● Public Key
● Curve● Generator (start)● Public x,y coordinate
● x509 extensions● Certificate Authority Signature
98
RSA ECC
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RSA Private Keys● Private Key
● Modulus (n) product of two prime numbers (p*q)● Public Exponent (e)● Private Exponent (d)● Prime1 (p)● Prime2 (q)
2008-05-29 Fun with Certificates
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ECC Private Keys● Private Key
● Private number (how many steps)● Public x,y coordinate● Public Generator (starting point)● Curve
102
RSA ECC
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Trust● Public Key Infrastructure (PKI)
● Certificate Authority (CA) i.e. notary● Intermediate Certificate● Client Certificate
● Web of Trust
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Public Key Infrastructure(PKI)
Web of Trust
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PKI● Why do we trust CAs?
● time consuming vetting process● regularly audited● $$$● bundled with product● certificate revocation
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Structure● Root CA
● self signed● Intermediate certificate● Server certificate
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Who provides the CA certificate, the client or the server?
The client.
...the intermediate certificate?
The server. (or it should)
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Getting Your Cert Signed● Internal Certificate Authority● Commercial Certificate Authority
● Be a reseller ($12/yr, $119/yr wildcard)● inCommon for .edu’s ($2k-$20k/yr)
● https://www.incommon.org/certificates/
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Semi-primes
41 * 43 = easy!1763
1739 = 47 * 37 difficult
1791904897 = 49943 * 35879
170122668341587273458646411386585043888873643113298660753168823105496218048396254258389541689798276387535036676575062116463749217204880781486238521463801806647717753763762209533452596443765433132839199250997874070119227832756249288919712152428105344288137338378592441098310151010596800002333954751873349228763 * 143685366445138003711595402594806625836106895764255994658099545498390517894693472991085893832864915801761970155763201096759761623694012072299292478856561357050062892354466628960025947611851554780658080196114743327960874693198902680721554877864174333388893106637708514607610834750473283277858418617695308935563 = 24444137941285645379511684911299365678423833046448779381238796084162536046797899019234205442218213499926991297229281024701278950648068677702332885730383357978977040184484121175079987603694398742376695650950853277837222494281038135867022877083226479856395867447419772143605903245226717018069307504429199930327344784767917383283267106133917174472280561457908186415882389738067587305825291144415722855157890883871648649466532813832921881732883942736314267482744271752456430649004239402313393638372879487394870568428620598721555293620836002747794896212943069775576590434653324242136440479444891894641015313209968513198569
hard, I need a computer!
Really hard, I need a super computer and a couple of millennia!
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Breaking semi-primes● Brute force● Sieve methods (slightly better)● Rainbow table
● What if we stored all 174 bit primes on micro-SD cards?
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100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001001101100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010100101100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010111111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000110000101100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000110101111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001100111111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001110001101100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010001111101100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010011111011100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010100010011100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010100111001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010111000011100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011000001111100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011011100001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011101000001100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011110000101.........................................................................
11972621413014756705924586149611790497021399392059337≅ 1.2*1052
23945242826029513411849172299223580994042798784118783≅ 2.4*1052
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001001101
111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101
Let's store every 174 bit prime number!
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How many?# primes = π(x) = x/ln(x)
x1 = 11972621413014756705924586149611790497021399392059392
x2 = 23945242826029513411849172299223580994042798784118783
π(x2) – π(x
1) ≅ 9.87*1049 primes
0.005 g 1pb microsd
solar mass 1.9891*1030 kg
174 bitsprime
1 byte8 bits
1 kb 1024 bytes
1 mb 1024 kb
1 gb 1024 mb
1 tb 1024 gb
1 pb 1024 tb
1 kg 1000 gx
x x x x
x x x x =
x9.87*1049 primes
4.8 solar masses ≅ ???
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Cert Lab
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Wrap-up● Cryptography● RSA overview● Explain why ECC came about● ECC deep dive● Safe Curves and Trust● Certs
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T
H
A
N
K
S
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