MHB How can we find the decrypted message?

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To decrypt the message sent from B to A using the ElGamal encryption system, the public key parameters are given as (p, g, y) = (53, 2, 27) and the encrypted message as (r, c) = (15, 34). The relationship between the public key and private key allows for the calculation of A's private key, x, from the equation 27 = 2^x (mod 53). Once x is determined, the original message can be found using the formula m = c / r^a, where a is derived from the private key. The discussion highlights the challenge of calculating discrete logarithms in practice, emphasizing the complexity of deriving the private key.
mathmari
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Hey! :o

Let's suppose that $A$ uses the encryption system of ElGamal with with public key $(p, g, y)=(53, 2, 27)$. $B$ sends to $A$ the encrypted message $(15, 34)$. Find the original message.

We have that $r=15, y=g^a, y^k=g^{ak}=r^a, c=m \cdot y^k=34$.

$$m=\frac{c}{r^a}=\frac{34}{15^a}$$

How can we continue?? (Wondering)

We are not given the private key of $A$, are we??

So, is the answer a function of $a$??
 
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mathmari said:
Hey! :o

Let's suppose that $A$ uses the encryption system of ElGamal with with public key $(p, g, y)=(53, 2, 27)$. $B$ sends to $A$ the encrypted message $(15, 34)$. Find the original message.

We have that $r=15, y=g^a, y^k=g^{ak}=r^a, c=m \cdot y^k=34$.

$$m=\frac{c}{r^a}=\frac{34}{15^a}$$

How can we continue?? (Wondering)

We are not given the private key of $A$, are we??

So, is the answer a function of $a$??

Hi mathmari,

The ElGamal cryptosystem as all other public key cryptosystems has a relationship between the public key and the private key. Note that,

\[27=y=g^x=2^x\mbox{ (mod }53)\]

We can calculate the value of $x$ which is the secret key. Although in practice the values will be larger and calculating the discrete logarithms are hard.
 
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