Math 336 - Prelim 1 - , 59 or 118 . 4. For m ≥ 2, show...

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Mathematics 336 Prelim 1 Solutions February 22, 2008 5 problems, 20 points each 1. Solve, if possible , each of the following. If not possible, state why. (a) Find an integer b so that [5] 37 · [ b ] 37 = [1] 37 . Solution: Since gcd(5 , 37) = 1 , there is a solution. By Bezout, we get 15(5) - 2(37) = 1 so [ b ] 37 = 15 . (b) Find an integer a so that [ a ] 420 · [91] 420 = [1] 420 . Solution: Since gcd(420 , 91) = 7 , there is no solution. 2. Compute the least nonnegative integer congruent to 4 119 modulo 7. Show your work. Solution: Since 4 2 2(mod 7) , 4 6 2 3 1(mod 7) . Then 4 119 = 4 19 · 6 4 5 4 5 4 2 · 4 2 · 4 2 · 2 · 4 2(mod 7) . 3. For which integers m 2 is it true that 3 5 5 3 (mod m )? Solution: 3 5 - 5 3 = 243 - 125 = 118 = 2 · 59 (both prime) so m = 2
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Unformatted text preview: , 59 or 118 . 4. For m ≥ 2, show that [ m-1] is always a unit in the ring Z /m Z . What is its order? Solution: [ m-1] m = [-1] m so [ m-1]-1 m = [ m-1] m . If m=2, its order is 1; otherwise it is 2. 5. Show that for any integer n , 21 divides n 25-n . Solution: Show both 3 and 7 divide n 25-n = n ( n 24-1) . By Fermat, either 3 | n or n 2 ≡ 1(mod 3) , in which case n 24 = ( n 2 ) 12 ≡ 1(mod 3) . Either way, 3 | n ( n 24-1) . Similarly, either 7 | n or n 6 ≡ 1(mod 7) , in which case n 24 = ( n 6 ) 4 ≡ 1(mod 7) . Again, we conclude 7 | n ( n 24-1) . 1...
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This note was uploaded on 02/24/2008 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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