# Using same modulus (n)

## Introduction

Alice and Bob encrypt the same message m (c1 for Alice and c2 for Bob).

The attacker intercepts the two encrypted messages c1 and c2 and the two public keys of Alice and Bob. Let's try to find the original message.

Alice : pubKey1(n, e1) and privKey1(n, d1)
Bob : pubKey2(n, e2) and privKey2(n, d2)

### Prerequisites

• We have c1 and c2 which are two encrypted messages from pubKey1 and pubKey2.
• pubKey1 and pubKey2 have the same modulus n (n1 = n2), however e1, e2, d1 and d2 can differ.
• c1 and c2 comes from the same cleartext m.

### Maths

Basic RSA :

$$c_{1} \equiv m^{e_{1}} [n]$$ $$c_{2} \equiv m^{e_{2}} [n]$$

Bezout's identity :

$$gcd(e_{1}, e_{2}) == 1$$ $$e_{1} * u + e_{2} * v = 1$$

Try to make something to the power of e1 * u + e2 * v :

$$c_{1}^u \equiv (m^{ e_{1} })^u \equiv m^{ {e_{1} } \times u} [n]$$

$$c_{2}^v \equiv (m^{ e_{2} })^v \equiv m^{ {e_{2} } \times v} [n]$$

$$m^{ {e_{1}} \times u } \times m^{ {e_{2} } \times v } \equiv m^{ {e_{1}} \times u + {e_{2}} \times v} \equiv m^1 \equiv m[n]$$

Conclusion :

$$c_{1}^u \times c_{2}^v \equiv m[n]$$

## Example

### Encryption using same modulus

Source code :

from Crypto.Util.number import bytes_to_long

m = bytes_to_long(b"S3CR3T!!!")
print("Plaintext (hex) =", m)

# Alice (1) and Bob (2)
n = 262680224351198943558562962102931091165978396557063906345939
e_1, e_2 = 65537, 34352

c_1 = pow(m, e_1, n)
c_2 = pow(m, e_2, n)
print("Encrypted (c1) =", c_1)
print("Encrypted (c2) =", c_2)


Output :

Plaintext (hex) = 1534773644617674989857
Encrypted (c1) = 188774664250657377040945189723460943665792254753522168336372
Encrypted (c2) = 183884636824805796552562064524263351724174474778356736863748


### Decryption with

• pubKey1(n, e1) and pubKey2(n, e2)
• c1 and c2

Source code :

from Crypto.Util.number import long_to_bytes

def bezout(a, b):
if b == 0:
return 1, 0
else:
q, r = a // b, a % b
x, y = bezout(b, r)
return y, x - q * y

u, v = bezout(e_1, e_2)
assert e_1 * u + e_2 * v == 1

cleartext = pow(c_1, u, n) * pow(c_2, v, n) % n
print("Cleartext :", long_to_bytes(cleartext).decode())


Output :

Cleartext : S3CR3T!!!