Outline. Cryptography. Bret Benesh. Math 331


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1 Outline 1 College of St. Benedict/St. John s University Department of Mathematics Math The internet is a lawless place, and people have access to all sorts of information. What is keeping people from stealing your credit card number when you purchase something online? Algebraic structures, that is what. is the study of sending and receiving secret messages. We will see how websites protect buyers through mathematics. References for this are of Gallian and Judson s free text at There are two ways to encrypt a message: via a private key encryption or via a public key encryption. In a private key encryption, one must know the complete code in order to either encrypt or decrypt the message. In public key encryption, one must know the complete code in order to decrypt, but only needs to know part of the code in order to encrypt.
2 We can translate our alphabet a, b,..., z into numbers by making a 01, b 02,... z 26 (perhaps a space translates to 00, so we have 27 symbols). We can encode the word bad by One code would be each letter n n + b for some fixed b; for example, f (n) n + 2 mod 27 is our encryption function. So bad is translated to , which is then encoded as dcf. To decode, you simply apply the additive inverse (subtract 2 module 27). If you know how to encode, it is really easy to figure out how to decode. We can encode the word bad by Let f (n) a n + b mod 27 for each letter, where gcd(a, 27) 1. For example, let f (n) 23 n + 2 mod 27. So bad is translated to , which is then encoded as uym. b 02 23(02) mod 27 a 01 23(01) mod 27 d 04 23(04) mod 27 To decode, you apply the inverse function f 1 (n) a 1 n a 1 b 20n 20(2) 20n 40 20n 13 (since (20)(23) (10)(46) (10)(19) (5)(38) (5)(11) 55 1 mod 27). For example 21 20(21) (15) mod 27 We can encode the word help by (our alphabet now goes a 00, b 01, etc). We will encode the letters in pairs for this code. So we will encode he together and lp together. Our encoding function will be f ( n) A n + b, where A is invertible with entries in Z26. Then the decoding function is f 1 ( x) A 1 x + A 1 b. [ ] [ ] For example, let f ( n) n [ ] [ ] [ ] Then f 1 ( x) x [ ] [ ] [ ] [ ] x + x +.Then [ ] [ ] h 07 e 04 [ ] [ ] 17 r 17 r [ ] [ ] l 11 p 15 [ ] [ ] 2 c 17 r [ [ ] [ ] ] [ ] So help encodes as rrcr. Decoding is similar. [ ] 2 2 [ ] 2 2 [ ] [ ] [ ] [ ]
3 A huge disadvantage to private key cryptography is that, in order to send coded messages, the sender must know the entire key. This is a problem for websites, since they would like everyone to be able to send codes (but they do not want everyone to know how to decrypt the messages). The solution is public key cryptography, where part, but not all, of the encode is made public. This system relies on hard problems to keep the public from knowing the entire coding function. RSA 1 Pick two large primes p, q. We will use p 3 and q Let n pq. In our example, n Let m (p 1)(q 1). In our example, m (2)(10) Pick positive integers at random until you find one that is relatively prime to m; call this integer E for encryption." We will use E 7. 5 Find a number D (for decryption") such that ED 1 mod m. We can use D 3, since DE 21 1 mod 20. RSA 1 Post both n and E to your website for the public to see. So we will post n 33 and E 7. 2 Someone can now send a message x to us now by letting y x E mod n and sending y to us. For example, someone might want to send us x 2 as a message. They would compute y 2 7 mod 33 (2 5 )(2 2 ) mod 33 ( 1)(4) mod 33 29, and send y 29 to us. 3 To decode, we simply compute y D mod n (D is secret, known only to you). For example, 29 3 mod mod 3 33(739) + 2 mod Try to solve the following problems. 1 Find an appropriate E and D if p 3 and q 5. Answer: e.g. E 3, D Using n 33, E 7, and D 3, encrypt x 6. Answer: y 6 7 mod Using n 33, E 7, and D 3, decrypt y 7. Answer: x 7 3 mod Suppose that you knew that n 35 and E 5 (which is always public information), and that you intercepted a code y 2. Figure out how you could break the code. Answer: x 2 5 mod 35 32
4 Why you might be worried. In the last problem with n 35 and E 5, there are two methods you probably thought of on how to crack the code: 1 (Brute force attack) For 0 i 34, check i E i 5 to see which i gives you y 2. 2 (Factoring attack) Factor n 35 5(7), and know that m (5 1)(7 1) 24. Find a number D such that ED 5D 1 mod 24. Why you should (mostly) not be worried. 1 (Brute force defense) We usually take p q, so n pq So you have to check about numbers by this method. If you had a computer that could check one trillion (10 12 ) of these per second, you would need seconds to do this. This should not worry you, since scientists think that the universe is approximately seconds old. 2 (Factoring defense) This relies on factoring huge numbers. This is something that is very hard for us to do right now. However, we are not sure that this is actually a difficult thing to do, we just cannot do it right now. So someone could come up with an easy way to factor and break RSA. Partial Proof for RSA Let n, m, E, and D be as above. Suppose we try to encrypt a message x with x relatively prime to n. Then the encrypted message is y x E, we can decrypt by y D (x E ) D x ED. We picked D so that ED 1 mod m, so there is an a such that ED am + 1. Then y D x ED x am+1 (x m ) a x. Partial Proof for RSA (continued) Since x is relatively prime to n, x is an element of (Cn {0}, ). This has order m (p 1)(q 1), so x m e 1. So y D (x m ) a x 1(x) mod n. (Note: We have not proved the result if (x, n) 1; this is a relatively easy number theory result using the Chinese Remainder Theorem).
5 Digital Signatures Suppose Alice is waiting to receive an encrypted message from Bob. Problem: Even if Bob sends Alice a properly encrypted message, how can Alice be sure that it was Bob who sent it? Digital Signatures (continued) Solution: Suppose that Alice has a public key (n, E) and a private key (n, D), and Bob has a public key (n, E ) and private key (n, D ). Here is what they can do: 1 Bob takes his message x, and encrypts it by using the key D normally used for decrypting, yielding x D. Only Bob could do this, since only Bob knows D. 2 Bob sends his new message x D to Alice using her encryption key (which everybody knows), so he sends her x D E. 3 Alice first decrypts using her private decryption key D, yielding (x D E ) D (x DE ) D x D mod n. 4 Alice then uses Bobs encryption key E to decrypt the message: (x D ) E x D E x mod n. So Alice gets the original message x, and knows that only Bob could have sent it. One more encryption scheme Instead of RSA, we could create an encoding scheme based on elliptic curves, which are planar equations of the form y 2 x 3 + ax + b (along with a point at infinity). These curves have a group structure associated with them. Instead of having the difficult problem being factoring large numbers," the difficult problem is solving the equation g n h where g and h are elements of a group G and n is an integer.
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