# Hw4_sols.pdf - Solution for Math470 Ziyi Zhou...

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Solution for Math470 Ziyi Zhou [email protected] BLOC605 Department of Mathematics, Texas A&M University September,2018 If you find any mistakes, please e-mail me. 1 Chapter 3 1.1 Section 3.4 Exercises3.14 (a). All we need to do is to show a 560 1 (mod 3), a 560 1 (mod 11), a 560 1 (mod 17). Because 3,11,17 are prime, by FLT: a 2 1 (mod 3) So ( a 2 ) 280 1 (mod 3), Similarly, ( a 10 ) 56 1 (mod 11), ( a 16 ) 35 1 (mod 17).. We know that a 561 - a = 3 i = 11 j = 13 k , for all a . Hence, a 561 - a = 3 * 11 * 17 l , for all a . (b). Here ( a 10 ) 7536 1 (mod 11) ( a 12 ) 6280 1 (mod 13) ( a 16 ) 4710 1 (mod 17) ( a 30 ) 2512 1 (mod 31) Obviously, we can use the method given before to show 75361 is a Carmichael number. Exercises3.15 (a). Here q = 69, we need to test a 2 i q (mod n ). Use Matlab: a ; n=1105; q=69; k=4; a=mod( aˆq , n ) ; F=0; i f mod(a , n)==1 return end f o r i =0:1:k - 1 i f mod(a , n)== - 1 return e l s e a=mod( a ˆ2 ,n ) ; end end 1
F=1; return Here. I think you can try by yourselves to get the answer. F = 0 means ’Test fails’, F = 1 means ’Composite’. Exercises3.17 (a). π (20) = 8 π (30) = 10 π (100) = 25 Exercises3.18 (a)(iii). π 1 ( X ) = 11(5 , 13 , 17 , 29 , 37 , 41 , 53 , 61 , 73 , 89 , 97) π 1 ( X ) = 13(3 , 7 , 11 , 19 , 23 , 32 , 43 , 47 , 59 , 67 , 71 , 79 , 83) 1.2 Section 3.5 Exercises3.22 (a). n = 1739,we find that: 2 3! - 1 63 (mod 1739) , gcd(2 3! - 1 , 1739) = 1