Problem 1 1/20/2011Practical Aspects of Modern Cryptography.

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Assignment #2 – Solutions
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Transcript of Problem 1 1/20/2011Practical Aspects of Modern Cryptography.

Assignment #2 – Solutions

Practical Aspects of Modern Cryptography

Problem 1 mod and mod (Note: mod is shorthand for mod .) mod mod

Four solutions

mod mod mod mod mod mod mod mod mod mod mod mod

1/20/2011

Practical Aspects of Modern Cryptography

Problem 2Given , select a

random , compute mod , and input and to the black box to produce output .

If mod , repeat above.Otherwise, compute gcd to produce a non-trivial factor of

.

1/20/2011

Practical Aspects of Modern Cryptography

Problem 2 – BonusRemove all powers of 2 from .Repeatedly use black box to split into prime powers .For each non-prime prime power,

try each of until an is found such that the root of is prime.

1/20/2011

Practical Aspects of Modern Cryptography1/20/2011

Problem 3Use Fermat’s Little Theorem and induction on k to prove

that

mod mod

for all primes and .

Practical Aspects of Modern Cryptography1/20/2011

Problem 3 (cont.)By induction on … Base case : mod mod mod Base case :

mod mod mod mod (by Fermat’s Little Theorem)

Practical Aspects of Modern Cryptography1/20/2011

Problem 3 (cont.)Inductive step:

Assume that mod mod .

Prove that mod mod .

Practical Aspects of Modern Cryptography1/20/2011

Problem 3 (cont.)mod mod mod mod mod (by inductive hypothesis) mod mod (by Fermat’s Little Theorem)

Practical Aspects of Modern Cryptography1/20/2011

Problem 4Show that for distinct primes and ,

mod mod mod mod

together imply that mod mod .

Practical Aspects of Modern Cryptography1/20/2011

Problem 4 mod mod mod mod is a multiple of .

Similarly mod mod is a multiple of .

Practical Aspects of Modern Cryptography1/20/2011

Problem 4Therefore, is a multiple of pq mod mod mod mod mod .

Practical Aspects of Modern Cryptography1/20/2011

Problem 5Put problems 3 and 4 together to prove that

mod mod For and distinct primes and .

Practical Aspects of Modern Cryptography1/20/2011

Problem 5 (cont.)Let and .

mod mod mod and

mod mod mod

By Problem #1, and then by Problem #2 mod mod .