week4 - 8/4/2010 SIT281 Introduction to Cryptography Week 4...

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8/4/2010 1 SIT281 Introduction to Cryptography Week 4 Objectives In this lecture we cover sections 3.5, 3.6, 3.7 and 3.9. The topics we study are: >Modular Exponentiation >Theorems of Fermat and Euler >Primitive roots >Finding square roots modulo n.
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8/4/2010 2 Modular exponentiation > As we mentioned earlier, the security behind many currently used cryptosystems is simply their size (recall the key bit-size charts from week 1). > So in cryptography, we like to ± Generate large numbers quickly in order to implement and Generate large numbers quickly in order to implement and secure systems, and ± Work out ways of reducing large number problems to smaller number problems when trying to decrypt (legitimately) or attack a system. Modular exponentiation- an example Suppose we want to evaluate a large power of 3 modulo 522. > We first evaluate the squares: 3 2 9 (mod 522) 3 4 (3 2 ) 2 81 (mod 522) 3 8 (3 4 ) 2 297 (mod 522) (3 ) 3 16 (3 8 ) 2 513 (mod 522) 3 32 (3 16 ) 2 81(mod 522) 3 64 (3 32 ) 2 81 2 297 (mod 522) and so on.
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8/4/2010 3 An example cont’d > We can now easily compute any power of 3 as follows. > As an example, consider 3 1234 . The exponent is not a power of 2, but we can write it as a sum of powers of 2 simply using its binary representation. > How do we get a binary representation of a number? We write it as a sum of powers of two. Writing a number as a sum of powers of 2: > First, find the highest power of 2 equal to or just below 1234: it is 1024. > Now look at the difference, and do the same thing: 1234 – 1024 = 210 and the highest power of 2 near (below) it is 128. > So 1234 = 1024 + 128 + (210 – 128) = 1024 + 128 + 82. > And 82 = 64 + 18 = 64 + 16 + 2. So: 1234 = 1024 + 128 + 64 + 16 + 2 = 2 10 + 2 7 + 2 6 + 2 4 +2.
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8/4/2010 4 Binary representation From 1234 = 2 10 +2 7 6 4 + 2 + 2 + 2 + 2 we obtain the corresponding binary representation of 1234 as 10011010010. Now back to our problem: So 3 1234 = 3 1024 * 3 128 * 3 64 * 3 16 *3 2 . We can compute the two highest powers: 3 128 (3 64 ) 2 297 2 513 (mod 522) 3 1024 (3 128 ) 8 513 8 (-9) 8 81 4 513 (mod 522) An example cont’d > Multiplying to obtain 3 1234 = 3 1024 * 3 128 * 3 64 * 3 16 2 , we have 3 1234 = (513*513)*297*(513*9) 81*297*(-81) 225*297 9 (mod 522). You will see more of these problems in the homework and prac.
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8/4/2010 5 Prime Numbers again We introduce this week a fundamental test for ‘non- primality’. How would you try to factor a 150 digit number? It may be a prime, and therefore not factorable at all. Since the standard factoring algorithm takes too long for most people, you would not be able to determine this in a reasonable amount of time. However, there are some ‘work-arounds’ that have been discovered. Prime Numbers again cont’d The result we prove below can sometimes tell you if a number is composite. Fermat’s Little Theorem. If p is a prime and p | a , then (yes, there is a Fermat’s Big Theorem but we don’t need it!) ). (mod 1 1 p a p Before giving a proof of this very useful result, we discuss some of the implications and do some examples.
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8/4/2010 6 Fermat’s Little Theorem (FLT) Suppose someone gives us a (large) number and we do not know if it’s prime or not.
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week4 - 8/4/2010 SIT281 Introduction to Cryptography Week 4...

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