Cryptography and security Assignment

Securing the Internet presents great challenges and research opportunities. Potential applications such as Internet voting, universally available medical records, and ubiquitous e-commerce are all being hindered because of serious security and privacy concerns. The epidemic of hacker attacks on personal computers and web sites only highlights the inherent vulnerability of the current computer and network infrastructure

Decryption process:

Plaintext need to be extracted from the message m.

13 = m^23 mod 143

n is product of 143

143 is divisible by 11 and 13

So assumption p=11 and q = 13
ϕ(pq)=(11-1)*(13-1) = 120

from modular multiplicative inverse, d = 47

therefore:
plain text = 13^47 mod 143 = 117

So plain text comes to be 117.

The correctness of decryption

To prove correctness of decryption can be done using inverse functions using Chinese Remainder Theorem,

It should prove inverse of decrypt as 13.

M^{ed} = M (mod p) M (mod q) =

M (mod pq) = M (mod n)

117^{23*47} =117(mod 143}

Encryption: c = m^{e} mod n, m < n

117^{23}mod 143  this is equal to 13 hence it is the correct decryption

2.

From input:

p=11, g=2

So p-1 = 10

As both 2 and 11 are co-prime

Now solve 2 ^ 10 mod 11

2^10 mod 11 does equal 1 so 2 has modulo order p-1 where p is a prime and

So hence derived that 2 is a primitive root of 11.

Proved.

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b

Given input q =11, n=2, Ya = 9.

Ya = (n^Xa) mod p = (2^Xa) mod 11 = 9

find the power of 2 has modulo so in order that 9 mod 11:

2^1 mod 11 = 2
2^2 mod 11 = 4
2^3 mod 11 = 8
2^4 mod 11 = 5
2^5 mod 11 = 10
2^6 mod 11 = 9

Hence Xa = 6

c.

To find the value of K:

K = (Yb ^ Xa) mod q

= 3^6 mod 11

= 3

Value of the secret key is 3.

See our Expert Advice Database Security Assignment

(6/100) Given values in input x = 6 mod 13 and x = 2 mod 17, find x.

Using Chinese Theorem:

In this case:

x ≡ 6 (mod 13) and

x ≡ 2 (mod 17)

17 x_1′ ≡ 1 (mod 13)

13 x_2′ ≡ 1 (mod 17)

To solve 17 x_1′ ≡ 1 (mod 13), note that

17 ≡ 4 (mod 13), hence just solve:

4 x_1′ ≡ 1 (mod 13)

=> x_1′ = 10, x_1 = 170

Solve: 13 x_2′ ≡ 1 (mod 17)

=> x_2′ =4, x_2 =52

Then your (unreduced) result is

x = Σ (a_i * x_i)

= 6 x_1 + 2 x_2

= 6*170 + 2*52 = 1020 + 104

= 1124

Finally we the result modulo for the 13*17:

x = 1124 ≡ 1124 – 5*221 (modulo 221) = 1

So x=19.

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