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# Assignment #3 Solutions - PowerPoint PPT Presentation

Assignment #3 Solutions. January 24, 2006. Problem #1. Use Fermat’s Little Theorem and induction on k to prove that x k ( p –1)+1 mod p = x mod p for all primes p and k ≥ 0. Answer #1. By induction on k … Base case k = 0 :

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### Assignment #3Solutions

January 24, 2006

Use Fermat’s Little Theorem and induction on k to prove that

xk(p–1)+1 mod p = x mod p

for all primes p and k ≥ 0.

Practical Aspects of Modern Cryptography

By induction on k …

Base case k = 0:

xk(p–1)+1 mod p = x0+1 mod p = x mod p

Base case k = 1:

xk(p–1)+1 mod p = x(p-1)+1 mod p

= xp mod p = x mod p

(by Fermat’s Little Theorem)

Practical Aspects of Modern Cryptography

Inductive step:

Assume that xk(p–1)+1 mod p = x mod p.

Prove that x(k+1)(p–1)+1 mod p = x mod p.

Practical Aspects of Modern Cryptography

x(k+1)(p–1)+1 mod p

= xk(p–1)+(p-1)+1 mod p

= xk(p–1)+1+(p-1) mod p

= xk(p–1)+1x(p-1) mod p

= xx(p-1) mod p (by inductive hypothesis)

= xp mod p

= xp mod p (by Fermat’s Little Theorem)

Practical Aspects of Modern Cryptography

Show that for distinct primes p and q,

x mod p = y mod p

x mod q = y mod q

together imply that

x mod pq = y mod pq.

Practical Aspects of Modern Cryptography

x mod p = y mod p

 (x mod p) – (y mod p) = 0

 (x – y) mod p = 0 (by first assignment)

 (x – y) is a multiple ofp.

Similarly x mod q = y mod q

 (x – y) is a multiple ofq.

Practical Aspects of Modern Cryptography

Therefore, (x – y) is a multiple ofpq

 (x – y) mod pq = 0

 (x mod pq) – (y mod pq) = 0

x mod pq = y mod pq.

Practical Aspects of Modern Cryptography

Put everything together to prove that

xK(p–1)(q-1)+1 mod pq = x mod pq

For K ≥ 0 and distinct primes p and q.

Practical Aspects of Modern Cryptography

Let k1=K(q–1) and k2=K(p–1).

xK(p–1)(q-1)+1 mod p = xk1(p–1)+1 mod p = x mod p

and

xK(p–1)(q-1)+1 mod q = xk1(q–1)+1 mod q = x mod q

By Problem #1, and then by Problem #2

xK(p–1)(q-1)+1 mod pq = x mod pq.

Practical Aspects of Modern Cryptography

E(x) = x43 mod 143

Find the inverse function

D(x) = xd mod 143.

Practical Aspects of Modern Cryptography

143 = 1113

We need to find d such that

43d mod (11–1)(13–1) = 1.

Use the Extended Euclidean Algorithm to find a solution to find x and y such that

120x + 43y = 1.

Practical Aspects of Modern Cryptography

Given A,B > 0, set x1=1, x2=0, y1=0, y2=1, a1=A, b1=B, i=1.

Repeat while bi>0: {i = i + 1;

qi = ai-1 div bi-1; bi = ai-1-qibi-1; ai = bi-1;

xi+1=xi-1-qixi; yi+1=yi-1-qiyi}.

For alli: Axi + Byi = ai. Finalai = gcd(A,B).

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Practical Aspects of Modern Cryptography

Digital Signature Algorithm

Public parameters: q = 11, p = 67, g = 9, y = 62

Private secret: x = 4

Message to be signed: M = 8

Selected random parameter: k = 2

Practical Aspects of Modern Cryptography

To sign a 160-bit message M,

• Generate a random integer k with 0 < k < q,

• Compute r = (gk mod p) mod q,

• Compute s = ((M+xr)/k) mod q.

The pair (r,s) is the signature on M.

Practical Aspects of Modern Cryptography

• r = (gk mod p) mod q

= (92 mod 67) mod 11

= (81 mod 67) mod 11 = 14 mod 11 = 3

• s = ((M+xr)/k) mod q

= ((8+43)/2) mod 11

= (20/2) mod 11 = 10 mod 11 = 10

The pair (3,10) is the signature on 8.

Practical Aspects of Modern Cryptography

A signature (r,s) on M is verified as follows:

• Compute w = 1/s mod q,

• Compute a = wM mod q,

• Compute b = wr mod q,

• Compute v = (gayb mod p) mod q.

Accept the signature only if v = r.

Practical Aspects of Modern Cryptography

• w = 1/s mod q = 1/10 mod 11 = 10

• a = wM mod q = 108 mod 11 = 3

• b = wr mod q = 103 mod 11 = 8

• v = (93628 mod 67) mod 11

= (5915 mod 67) mod 11

= 14 mod 11 = 3

v = 3 andr = 3 so the signature is validated.

Practical Aspects of Modern Cryptography