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### Assignment #3Solutions

January 24, 2006

Problem #1

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

Answer #1

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

Answer #1 (cont.)

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

Answer #1 (cont.)

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

Problem #2

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

Answer #2

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

Answer #2 (cont.)

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

Problem #3

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

Answer #3

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

Problem #4

E(x) = x43 mod 143

Find the inverse function

D(x) = xd mod 143.

Practical Aspects of Modern Cryptography

Answer #4

143 = 1113

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

Extended Euclidean Algorithm

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

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Answer #4 (cont.)

Practical Aspects of Modern Cryptography

Problem #5

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

The Digital Signature Algorithm

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

Answer #5

- 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+43)/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

The Digital Signature Algorithm

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

Answer #5 (cont.)

- w = 1/s mod q = 1/10 mod 11 = 10
- a = wM mod q = 108 mod 11 = 3
- b = wr mod q = 103 mod 11 = 8
- v = (93628 mod 67) mod 11

= (5915 mod 67) mod 11

= 14 mod 11 = 3

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

Practical Aspects of Modern Cryptography

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