MHB How can we find the decrypted message?

[mathmari]

Hey!

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 $a$??

 

[Sudharaka]

Hey!

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.