HW5_solns - MATH 3360 Sketch of solutions for Assignment 5 9C.49 Since(18 = 6 we have 56 = 5 55 1(mod 18 Answer 55 9C.50[graded out of 2 points f

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MATH 3360 Sketch of solutions for Assignment 5 9C.49 Since ϕ (18) = 6, we have 5 6 = 5 · 5 5 1(mod 18). Answer: 5 5 . 9C.50 [graded out of 2 points] f 1(mod ϕ ( m )) f = ( m ) + 1 for some k Z . Thus, a f = a ( m )+1 = ( a ϕ ( m ) ) k · a 1 · a (mod m ) by Euler. 9C.62 [graded out of 5 points] We are given that the order of 2 is q - 1, so 2 q - 1 1(mod q ). We also have ( q, 2) = 1, so 2 is a unit and by Euler 2 ϕ ( q ) 1(mod q ). Since q - 1 is the order of 2, we must have q - 1 ϕ ( q ). By definition of ϕ , we must also have ϕ ( q ) q - 1, so ϕ ( q ) = q - 1 q is prime. Note that 2 10 1(mod 341). 10A.7 [graded out of 2 points] m = pq , ϕ ( m ) = ( p - 1)( q - 1) = pq - p - q + 1 = m - p - m/p + 1 ( m ) = pm - p 2 - m + p p 2 + ( ϕ ( m ) - m - 1) p + m = 0. Also note that m and ϕ ( m ) are both symmetric in p and q . Thus, a quadratic equation x 2 + ( ϕ ( m ) - m - 1) x + m = 0, has both p and q as its solution. 11B.11
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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HW5_solns - MATH 3360 Sketch of solutions for Assignment 5 9C.49 Since(18 = 6 we have 56 = 5 55 1(mod 18 Answer 55 9C.50[graded out of 2 points f

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