Math 115 - Spring 1998 - Ribet - Final

Math 115 - Spring 1998 - Ribet - Final - theory deduce the...

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Math 115 Professor K. A. Ribet Final Exam May 18, 1998 The numbers 257 and 661 are prime. 1 (6 points) . Find a positive integer n such that n/ 3 is a perfect cube, n/ 4 is a perfect fourth power, and n/ 5 is a perfect fifth power. 2 (5 points) . Prove that there are no whole number solutions to the equation x 2 - 15 y 2 = 31. 3 (5 points) . Find the number of solutions to the congruence x 2 9 mod 2 3 · 11 2 . 4 (7 points) . Which positive integers m have the property that there a primitive root mod m ? (Summarize what we know about this question, and why we know it. Your answer should be clear enough that one could use it to decide immediately if there is a primitive root modulo (257) 2 , 4 · 661, 257 · 661, . . . .) 5 (6 points) . Fermat showed that 2 37 - 1 is composite by finding a prime factor p of 2 37 - 1 which lies between 200 and 300. Using your knowledge of number
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Unformatted text preview: theory, deduce the value of p . 6 (7 points) . The continued fraction expansion of √ 5 is h 2 , 4 , 4 , . . . i . If h 2 , 4 , 4 , . . . , 4 | {z } 99 4 s i = h/k (in lowest terms), calculate h 2-5 k 2 . 7 (5 points) . Prove that there are an infinite number of primes congruent to 3 mod 4. 8 (6 points) . Suppose that p = a 2 + b 2 , where p is an odd prime number and a is odd. Show that ± a p ² = +1. (Use the Jacobi symbol.) 9 (8 points) . Let a and b be positive integers. Show that φ ( ab ) φ (gcd( a, b )) = φ ( a ) φ ( b ) gcd( a, b ) , φ = Euler φ-function . (Example: If a = 12 and b = 8, the equation reads 32 · 2 = 4 · 4 · 4.) 10 (5 points) . Find all solutions in integers y and z to the equation 6 2 + y 2 = z 2 ....
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This note was uploaded on 10/31/2009 for the course STAT 131A taught by Professor Isber during the Spring '08 term at Berkeley.

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