MATHEMATIC
Problem Set 4. Number theory

# Problem Set 4. Number theory - Problem Seminar Fall 2013...

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Problem Seminar. Fall 2013. Problem Set 4. Number theory. Classical results. 1. Euler. For a positive integer n and any integer a relatively prime to n one has a φ ( n ) 1 (mod n ) , where φ ( n ) is the number of positive integers between 1 and n relatively prime to n . 2. Wilson. ( p - 1)! ≡ - 1(mod p ) for any prime p . 3. Chinese Remainder theorem. Let m 1 , m 2 , . . . , m k be pairwise positive integers greater than 1 , such that gcd( m i , m j ) = 1 for i 6 = j . Then for any integers a 1 , a 2 , . . . , a k the system of congruences x a 1 (mod m 1 ) , x a 2 (mod m 2 ) , . . . x a k (mod m k ) . has solutions, and any two such solutions are congruent modulo m = m 1 m 2 . . . m k . Problems. 1. Put 1989. A1. How many primes among the positive integers, written in the usual base 10 , are such that their digits are alternating 1 s and 0 s, beginning and ending with 1 ? 2. Put 1993. B1. Find the smallest positive integer n such that for every integer m , with 0 < m < 1993 , there exists an integer k for which m 1993 < k n < m + 1 1994 . 3. GA 735. Solve in positive integers the equation x x + y = y y - x . 4. IMO 1970 . Prove that there are no positive integers n such that the set

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Unformatted text preview: + 1 ,n + 2 ,...,n + 6 } can be divided into two sets with the product of elements in one set equal to the product of elements in the other set. 5. Put 1983. A3. Let p be an odd prime, and let F ( n ) = 1 + 2 n + 3 n 2 + ... + ( p-1) n p-2 . Prove that if a and b are not congruent modulo P then F ( a ) and F ( b ) are not congruent modulo p . 6. IMO 2002. The positive divisors of an integer n > 1 are 1 = d 1 < d 2 < ... < d k = n . Let s = d 1 d 2 + d 2 d 3 + ... + d k-1 d k . Prove that s < n 2 and ﬁnd all n for which s divides n 2 . 7. Put 2001. A5. Prove that there are unique positive integers a,n such that a n +1-( a +1) n = 2001 . 8. Put 1996. A6. The sequence a n is deﬁned by a 1 = 1 ,a 2 = 2 ,a 3 = 24 , and, for n ≥ 4 , a n = 6 a 2 n-1 a n-3-8 a n-1 a 2 n-2 a n-2 a n-3 Show that, for all n , a n is an integer multiple of n ....
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