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.
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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
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
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To sign a 160-bit message M,
The pair (r,s) is the signature on M.
Practical Aspects of Modern Cryptography
= (92 mod 67) mod 11
= (81 mod 67) mod 11 = 14 mod 11 = 3
= ((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
A signature (r,s) on M is verified as follows:
Accept the signature only if v = r.
Practical Aspects of Modern Cryptography
= (5915 mod 67) mod 11
= 14 mod 11 = 3
v = 3 andr = 3 so the signature is validated.
Practical Aspects of Modern Cryptography