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examples/rsa_problem.py

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+#!/usr/bin/env python3
+
+from Cryptotools.Numbers.coprime import phi
+from Cryptotools.Utils.utils import gcd
+
+def generate_keys():
+    p = 7853
+    q = 7919
+    n = p * q
+    e  = 65536 # It's public value, must be coprime with phi n
+    
+    print(n)
+    #phin = phi(n)
+    phin = (p - 1) * (q - 1)
+    print(phin)
+    
+    for _ in range(2, phin):
+        if gcd(phin, e) == 1:
+            break
+        e += 1
+    
+    print(e)
+    plaintext = 'A'
+    ciphertext = pow(ord(plaintext), e, n)
+    print(f"Ciphertext: {ciphertext}")
+
+# Now, we can test
+
+# To resolve that formula: C = x ** e mod n
+# Where C is the ciphertext, and C, e and n are known (public values)
+# First, we need to find the reverse modular of phi(n) or carmi(n)
+# z = e -1 mod phi(n)
+# After that, we have our decryption key, we can resolve x
+# x = C ** z mod n
+# The RSA Problem is to decrypt with the public-key
+# We just need to find the decryption key with the public-key and the modulus
+
+# First, we need to find phi(n)
+# phin = phi(n) # I computed here, result = 62172136
+
+n = 62187907
+phin = 62172136
+e = 65537
+ciphertext = 38605768
+# print(phin)
+
+# Find the reverse modular
+d = pow(e, -1, phin)
+# print(d)
+plaintext = pow(ciphertext, d, n)
+print(chr(plaintext))
+