assignment #3 solutions

Assignment #3Solutions

January 24, 2006

January 24, 2006Practical Aspects of Modern Cryptography

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.


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)


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.


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 = x x(p-1) mod p (by inductive hypothesis) = xp mod p = xp mod p (by Fermat’s Little Theorem)


Problem #2

Show that for distinct primes p and q,x mod p = y mod px mod q = y mod q

together imply thatx mod pq = y mod pq.


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 of p.

Similarly x mod q = y mod q (x – y) is a multiple of q.


Answer #2 (cont.)

Therefore, (x – y) is a multiple of pq (x – y) mod pq = 0 (x mod pq) – (y mod pq) = 0 x mod pq = y mod pq.


Problem #3

Put everything together to prove thatxK(p–1)(q-1)+1 mod pq = x mod pq

For K ≥ 0 and distinct primes p and q.


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 pand

xK(p–1)(q-1)+1 mod q = xk1(q–1)+1 mod q = x mod qBy Problem #1, and then by Problem #2

xK(p–1)(q-1)+1 mod pq = x mod pq.


Problem #4

E(x) = x43 mod 143Find the inverse function

D(x) = xd mod 143.


Answer #4

143 = 1113We 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.


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 all i: Axi + Byi = ai. Final ai = gcd(A,B).


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 00 1


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 2

1 -2


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 23 34 9 1 -2

-1 3


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 23 34 9 1 -2 14 9 7 -1 3 3

4 -11


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 23 34 9 1 -2 14 9 7 -1 3 35 7 2 4 -11 1

-5 14


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 23 34 9 1 -2 14 9 7 -1 3 35 7 2 4 -11 16 2 1 -5 14 3

19 -53


Answer #4 (cont.)

i ai bi xi yi qi

1 120 43 1 02 43 34 0 1 23 34 9 1 -2 14 9 7 -1 3 35 7 2 4 -11 16 2 1 -5 14 37 1 0 19 -53 2

-43 120


Problem #5

Digital Signature Algorithm

Public parameters: q = 11, p = 67, g = 9, y = 62Private secret: x = 4Message to be signed: M = 8Selected random parameter: k = 2


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.


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


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.


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 and r = 3 so the signature is validated.

assignment #3 solutions

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