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135_9_-_Solutions

# 135_9_-_Solutions - Assignment 9 Solutions 1(a Given n = pq...

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Assignment 9 – Solutions 1. (a) Given n = pq , p > q , and φ ( n ) = ( p - 1)( q - 1) prove that p + q = n - φ ( n ) + 1 and p - q = p ( p + q ) 2 - 4 n. Solution: For the first part, n - φ ( n ) + 1 = pq - ( p - 1)( q - 1) + 1 = pq - ( pq - p - q + 1) + 1 = p + q. And for the second part, ( p + q ) 2 - 4 n = p 2 + 2 pq + q 2 - 4 pq = p 2 - 2 pq + q 2 = ( p - q ) 2 . Since ( p - q ) > 0 , we have p ( p + q ) 2 - 4 n = p - q . (b) The integer n = 71531 is the product of two primes p and q , and the Euler phi function φ ( n ) = ( p - 1)( q - 1) = 70992. Determine the prime factors p and q . Solution: We have n = pq = 71531 φ ( n ) = ( p - 1)( q - 1) = 70992 Since ( p - 1)( q - 1) = pq - ( p + q ) + 1 , we have 70992 = 71531 - ( p + q ) + 1 . Thus, p + q = 540 , pq = 71531 . Substituting p = 540 - q gives q 2 - 540 q + 71531 = 0 and thus q = 270 - p 270 2 - 71531 = 270 - 37 = 233 and p = 307. Check: (233)(307) = 71531 and (232)(306) = 70992. 1

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2. Use Fermat’s Little Theorem and the Square and Multiply Algorithm to prove that the integer 2479 is not prime (without testing each prime p 2479 to see if is a factor). Solution: We calculate 2 2478 (mod 2479). We make a list of powers of 2 modulo 2479. k 2 k 1 2 2 4 4 16 8 256 16 1082 32 636 k 2 k 64 419 128 2031 256 2384 512 1588 1024 601 2048 1746 Note that 2478 = 2048 + 256 + 128 + 32 + 8 + 4 + 2 so we have 2 2478 2 2048 · 2 256 · 2 128 · 2 32 · 2 8 · 2 4 · 2 2
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