diff --git a/misc/Proj RSA2_ Cracking a Short RSA Key (15 pts.).html b/misc/Proj RSA2_ Cracking a Short RSA Key (15 pts.).html new file mode 100644 index 0000000..2e2cc0d --- /dev/null +++ b/misc/Proj RSA2_ Cracking a Short RSA Key (15 pts.).html @@ -0,0 +1,310 @@ + + + +Proj RSA2: Cracking a Short RSA Key (15 pts.) + + +

Proj RSA2: Cracking a Short RSA Key (15 pts.)

+What you need: + +

Purpose

+To break into RSA encryption without prior knowledge of the +private key. This is only possible for small RSA keys, +which is why RSA keys should be long for security. +

+

Summary

+Here's a diagram from the textbook showing the +RSA calculations. +
+

Problem Statement

+Meghan's public key is (10142789312725007, 5). +Find her private key. +

1. Factoring n

+

Finding the Square Root of n

+n = 10142789312725007. This is the product of +two prime numbers, p and q. +

+How large are p and q? Well, they can't both be +larger than the square root of n, or they'd be larger +than n when multiplied together. +

+Start Python in interactive mode. On a Mac +or Linux box, you can do that by +typing this command into a Terminal window: +

+python +
+Execute these commands: +
import math
+n = 10142789312725007
+print math.sqrt(n)
+
+The square root prints out, +as shown below. +
+

Displaying More Decimal Places

+It's not clear from that output whether the result +is an integer, or just rounded off to one decimal +place. To see more decimal places, we'll use +the repr() function. +

+Execute this command: +

print repr(math.sqrt(n))
+
+Now more decimal places appear, +as shown below. +
+

Testing 20 Candidates

+So one of the prime factors must be a prime number +less than 100711415. All we have to do is try dividing +n by odd numbers starting at 100711413 and going down until +we get an integer result. (We don't need to test 100711415 because +it's divisible by 5 and therefore not a prime number.) +

+A good way to do this is to calculate n mod c, where c is a candidate. If c is a factor of n, the result will be zero. +

+We can test the first 20 candidates with a for loop. +

+Execute these commands: +

c = 100711413
+for i in range(c, c-40, -2):
+  print i, n%i
+
+Press Enter twice after the last command to terminate the loop. +

+The third candidate is the winner, with a remainder of zero, +as shown below. +

+

Calculating q

+We now know p and we can calculate q. +

+Execute these commands: +

p = 100711409
+q = n / p
+print p, q, n, p*q, n - p*q
+
+The calculation worked, so the last value is zero, +as shown below. +
+

2. Compute phin = (p-1) * (q-1)

+Execute these commands: +
phin = (p-1) * (q-1)
+print p, q, n, phin
+
+The parameters print out, as shown below. +
+

3. Compute Private Key d

+We need to find a d with this +property: +
+ (d * e) mod phin = 1 +
+We know that e = 5 from the Problem Statement. +

+It's not obvious how to find d, but +there's +a simple way to do it in Python, using the +"gmpy> library. +

+Open a new Terminal window, not the one +running Python, and execute this command to +download and install a few dependencies and gmpy: +

brew install gmp mpfr mpc
+pip install gmpy
+
+

+In the Terminal window running python, +execute these commands. +

e = 5
+import gmpy
+d = gmpy.invert(e, phin)
+print d, e, d*e %phin
+
+We get the value of d, and, to verify it, +we see that d*e %phin is indeed 1, +as shown below. +
+

4. Encrypting a Message

+

Encoding the Message as a Number

+Cueball wants to send Meghan this message: +
Hi!
+
+We can only send numbers. +Let's convert that message to three bytes of ASCII +and then interpret it as a 24-bit binary value. +

+In the Terminal window running python, +execute these commands. +

x1 = ord('H')
+x2 = ord('i')
+x3 = ord('!')
+x = x1*256*256 + x2*256 + x3
+print x
+
+We get the value of x, +as shown below. +
+

Encrypting the Message with the Public Key

+A public key contains two numbers: n +and e. To encrypt a message x, +use this formula: +
+

+Execute these commands: +

y = x ** e % n
+print y
+
+The encrypted message appears, +as shown below. +
+

5. Decrypting a Message

+To decrypt a message, use this formula: +
+Execute these commands: +
xx = y ** d % n
+print xx
+
+Python crashes, +as shown below. It cannot handle such large numbers. +
+To compute such a number, we must use the pow() function. +

+Execute these commands to restart python and restore +all the values we calculated previously: +

python
+n = 10142789312725007
+p = 100711409
+q = 100711423
+phin = (p-1) * (q-1)
+e = 5
+import gmpy
+d = gmpy.invert(e, phin)
+x1 = ord('H')
+x2 = ord('i')
+x3 = ord('!')
+x = x1*256*256 + x2*256 + x3
+y = x ** e % n
+
+Your screen should look like the image +below. +
+Let's try that decryption again with the pow() +function. Execute these commands: +
xx = pow(y, d, n)
+print xx
+
+Now it works, showing our original message in +numerical form. +
+

Converting the Message to Readable Text

+To convert that number back to letters, +execute these commands: +
xx1 = xx / (256*256)
+xx2 = (xx - 256*256*xx1) / 256
+xx3 = xx - 256*256*xx1 - 256*xx2
+msg = chr(xx1) + chr(xx2) + chr(xx3)
+print xx1, xx2, xx3, msg
+
+Now it works, showing the original message in +readable form, as shown below. +
+
+

Challenge 2a: Encrypt "OBEY!"

+Using the same system and keys, encrypt this message: +
OBEY!
+
+Hint 1: The message, converted to a number, is 12 digits long and ends in 41. +

+Hint 2: The encrypted message is 16 digits long and ends in 81. +

+Use the form +below to put your name on the + +WINNERS PAGE. +

+
+
+ + + + + + +
Your Name (without spaces):
Encrypted message:
+
+
+
+
+
+

Challenge 2b: Message to Cueball

+Cueball's public key is: +
(111036975342601848755221, 13)
+
+Meghan sends this message to Cueball. Decrypt it. +
80564890594461648564443
+
+Use the form +below to put your name on the + +WINNERS PAGE. +
+
+
+ + + + + + +
Your Name (without spaces):
Cleartext Message:
+
+
+
+
+
+

Challenge 3: Message to Rob

+Rob public key is: +
(1234592592962967901296297037045679133590224789902207663928489902170626521926687, 5557)
+
+Meghan sends this message to Rob. Decrypt it. +
272495530567010327943798078794037733865151017104532777645776936985235709526002
+
+Hint: +make square root calculations more precise. +

+Use the form +below to put your name on the + +WINNERS PAGE. +

+
+
+ + + + + + +
Your Name (without spaces):
Cleartext Message:
+
+
+
+
+
+

Sources

+ +Cracking short RSA keys +

+ +Prime Numbers Generator and Checker +

+Arbitrary precision of square roots +

+Posted 3-31-16 by Sam Bowne
+Winners pages added 8-6-16
+Attack server name updated 4-4-17
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