james c. bradas, ph.d

James C. Bradas, Ph.D.

Engineering & Analysis Operation-Part2Engineering & Analysis Operation-Part2

18 June 2009

Public Key Encryption

Alice

Private Key

Public Key

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( n , e )Public Key

( n , d )Private Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( n , e )Public Key

( n , d )Private Key

For this example, we’ll use small prime numbers – the principal is exactly the same….

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( n , e )Public Key

( n , d )Private Key

let p = 11, q = 3

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( 33 , e )Public Key

( 33 , d )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( 33 , e )Public Key

( 33 , d )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

20210)13)(111( 4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( 33 , 3 )Public Key

( 33 , d )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

20210)13)(111( 4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

( 33 , 3 )Public Key

( 33 , d )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

20210)13)(111( 4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

120

121

20

173 7

20

13 )20(mod13

d

kd

d

Check

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

( 33 , 3 )Public Key

( 33 , 7 )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

20210)13)(111(

120

121

20

173 7

20

13 )20(mod13

d

kd

d

Check

Alice

Private Key

Public Key

Let’s Try An Example to See How This WorksRSA Public Key Encryption Scheme

1. Select two large prime numbers, p & q

)1)(1( qp

2. Compute their product – the “modulus” n:

3. Compute the “totient”

4. Choose e, 1 < e < such that greatest common divisor (gcd) ( e , = 1

e is the “public key exponent”

( Common choices are e = 3, 17 & 65537 )

5. Compute d such that )(mod1 ed

( 33 , 3 )Public Key

( 33 , 7 )Private Key

let p = 11, q = 3

n = 11 x 3 = 33

20210)13)(111(

120

121

20

173 7

20

13 )20(mod13

d

kd

d

Check

)(modnmc eTherefore, we want to encrypt m = 7

( 33 , 3 )Public Key

n e

Suppose Bob Wants to Send Alice the Letter “Z”

Let’s say that in the agreed-upon reversible padding scheme, “Z” equals the number 7.Alice Bob

“Z”

)33(mod73c

n

yINTny is a solution to )(modnyx

Recall

)33(mod713

13

13

330343

1033343

33

34333343

)33(mod343

)33(mod7

3

3

c

INT

c

c

So….

...) 393939.10(10

... 393939.1033

343

INT

so

1033

330

33

34313

Check

)(modnmc eAlice Bob

Bob sends c = 13 to Alice

“Z”

13

)33(mod73

c

c

After Computing

Alice Bob

( 33 , 7 )Private Key

nd

)(modncm d

7

,

7

)33(mod4693)33(mod131919

)33(mod13)33(mod2197)33(mod2197

)33(mod13)33(mod13)33(mod13

)33(mod131313

33by divided 13 ofRemainder

)33(mod13

33

33

7

7

m

Therefore

m

Alice Receives “13”

Alice Bob

( 33 , 7 )Private Key

nd

)(modncm d

7

,

7

)33(mod4693)33(mod131919

)33(mod13)33(mod2197)33(mod2197

)33(mod13)33(mod13)33(mod13

)33(mod131313

33by divided 13 ofRemainder

)33(mod13

33

33

7

7

m

Therefore

m

Alice Receives “13”

ReversiblePaddingScheme

ReversiblePaddingScheme

Alice Bob

( 33 , 7 )Private Key

nd

)(modncm d

7

,

7

)33(mod4693)33(mod131919

)33(mod13)33(mod2197)33(mod2197

)33(mod13)33(mod13)33(mod13

)33(mod131313

33by divided 13 ofRemainder

)33(mod13

33

33

7

7

m

Therefore

m

ReversiblePaddingScheme

ReversiblePaddingScheme “Z”

“Z”

Alice Bob

Bob’s Message)(modnmc e

M m

M

)(modncm d

Encrypt

Decrypt

Private Key

Public Key

How does this Work?

So What’s Going On Here?

Some more properties we need to know

If )(mod11 nba )(mod22 nba and

))(mod()( 2121 nbbaa

))(mod()( 2121 nbbaa aaa 21

bbb 21)(mod22 nba

)(mod'' nba

onsoand

nba

nbbaa

nbbaa

)(mod

)(mod

)(mod''

33

22

let

)(mod

)(mod

nba

Then

nba

If

dd

)(mod)(mod

)(mod)(mod

21

21

nbnb

nbnb Then

)(modnba

… and more properties

Fermat’s Little Theorem: If p is a prime number, then for ANY integer a,

will be evenly divisible by p.aa p

)(mod paa p

)(mod11 pa p

or

)(mod1

,

)( na

then

ntocoprimeisaIf

n

Euler’s Theorem (An Extension of Fermat’s Little Theorem)

gcd(a,n)=1

Φ(n) = (p-1)(q-1)is Euler’s “Totient”

Here’s What I Want to Prove

)(modnmc e

)(modncm d

I can recover m via

pqn)1)(1( qp

)(mod1 ed

1.

2.

3.

Given:

Then:

If:

)(mod

)(mod

nba

nba

dd

)(modnmc e

Let’s raise c to the d power and use

)(mod

)(mod)(

nmc

nmcedd

ded

Here’s The Details

)(mod

nba

FormtheHasWhich

Start With:

)(mod

)(mod)(

nmc

nmcedd

ded

Here’s The Details

)(mod1 Now, ed

ked 1 means,Which

)(modnmc e)(mod

nba

FormtheHasWhich

)(mod

)(mod)(

nmc

nmcedd

ded

ked 1 So,

)(mod)(mod 1 nmnmc kedd kk mmm1

Here’s The Details

)(mod1 Now, ed

ked 1 means,Which

)(modnmc e)(mod

nba

FormtheHasWhich

)(mod

)(mod)(

nmc

nmcedd

ded

)(mod1 Now, ed

ked 1 means,Which

ked 1 So,

)(mod)(mod)(mod 1 nmmnmnmc kkedd kk mmm1

Here’s The Details

)(modnmc e)(mod

nba

FormtheHasWhich

)(mod

)(mod)(

nmc

nmcedd

ded

)(mod1 Now, ed

ked 1 means,Which

ked 1 So,

)(mod)(mod)(mod 1 nmmnmnmc kkedd kk mmm1

)(mod1)( nm n

Now, recall that

Here’s The Details

Euler’s Theorem

)(modnmc e)(mod

nba

FormtheHasWhich

)(mod

)(mod

)(mod1)(mod

)(mod1)(mod))(mod1)((mod

))(mod)((mod)(mod)(mod

nmc

nm

nnm

nnmnnm

nmnmnmmnmc

d

kk

kkedd

Here’s The Details(cont’d)

)(modncm d

Here’s The Details(cont’d)

)(mod

)(mod

)(mod1)(mod

)(mod1)(mod))(mod1)((mod

))(mod)((mod)(mod)(mod

nmc

nm

nnm

nnmnnm

nmnmnmmnmc

d

kk

kkedd

Which Can Be Written

)(modncm d

Here’s The Details(cont’d)

Which is What I Wanted to Prove

)(mod

)(mod

)(mod1)(mod

)(mod1)(mod))(mod1)((mod

))(mod)((mod)(mod)(mod

nmc

nm

nnm

nnmnnm

nmnmnmmnmc

d

kk

kkedd

)(modncm d

knba

nba

)(mod

knbaab

nab

)(

)(mod

Except for the change in sign, the two terms are equivalent.

This is because:

Here’s The Details(cont’d)

Which is What I Wanted to Prove

)(mod

)(mod

)(mod1)(mod

)(mod1)(mod))(mod1)((mod

))(mod)((mod)(mod)(mod

nmc

nm

nnm

nnmnnm

nmnmnmmnmc

d

kk

kkedd

So, Where Does RSA Encryption Stand?

• For now, RSA PKE is still secure

• In 1991, RSA Laboratories published 54 large semiprimes (numbers

with exactly two prime factors) and issued cash prizes for successful

factorization.

• According to Wikepedia, 12 of the 54 listed numbers had been factored

by March 2008

• The RSA challenge officially ended in 2007

• Fastest Published Integer Factorization Algorithms:

General Number Field Sieve

Quadratic Sieve

• Development of a large Q-Bit Quantum Computer MIGHT make RSA

vulnerable, although this is not certain

• Fundamental breakthroughs in Number Theory (such as solving the

Riemann Hypothesis) still required before RSA becomes vulnerable

"The whole of e-commerce depends on prime numbers. I have

described the primes as atoms: what mathematicians are

missing is a kind of mathematical prime spectrometer.

Chemists have a machine that, if you give it a molecule, will tell

you the atoms that it is built from. Mathematicians haven't

invented a mathematical version of this. That is what we are

after. If the Riemann hypothesis is true, it won't produce a

prime number spectrometer. But the proof should give us more

understanding of how the primes work, and therefore the proof

might be translated into something that might produce this

prime spectrometer. If it does, it will bring the whole of e-

commerce to its knees, overnight. So there are very big

implications." - Marcus du Sautoy (“The Music of the Primes”)

Questions