hw6 - p. 193 #11. Suppose that a nd W e cannot have = ,...

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p. 193 #11. Suppose that and We cannot have = , since otherwise by unique factorization in the integers. Without loss of generality assume that = , the case of being similar. Then we can quickly compute by using the Euclidean algorithm on and . However, once we know one of the proper factors of each of the n's, then by division we can quickly find the other factor. Knowing both factors of n, we can find in both cases. Once we know , we can find the modular inverse of the public encryption exponent e in both the cases, yielding the decryption exponents d in both cases. The RSA systems are broken. p. 193 #14. Since = , it follows that pq | (x + a)(x - a). Hence p | (x+a)(x-a), and in particular, the prime p divides one of the two factors. However, pq does not divide x + a or x - a. Therefore p divides only one factor while q divides only the other. Without loss of generality, assume that x + a = and x - a = . Hence, = = , or . We would like to find k and j. But since , we know that there exist integers s and t for which . Taking and . we get find solutions for k and j such that . We know that = since Now let x be the common value of these two expressions: x = = . We now show that such an x has the desired property. We see that = ( Hence, | . Using the other form of x, we see that so | Since p and q are distinct primes, it follows that | Hence, = For example, with p = 71 and q = 73, then (36)(71) - (35)(73) = 1, so, taking a = 7,
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(3.1) (2.1) restart: (14*36)*71-7; %^2 mod(71*73); 35777 49 p. 195 #22. a) Since
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hw6 - p. 193 #11. Suppose that a nd W e cannot have = ,...

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