MATH 145 Algebra, Solutions to Assignment 4

Transcription

1 MATH 145 Algebra, Solutions to Assignment 4 1: a Let a 975 and b161 Find d gcda, b and find s, t Z such that as + bt d Solution: The Euclidean Algorithm gives , , , , and so we have d gcda, b 13 Bac-Substitution then gives the sequence s 0 1, s 1, s 5, s 3 17, s 4 and so we can tae s and t 17 to get as + bt d b Solve the linear diophantine equation 33x + 765y 1 Solution: Let a 33, b 765 and c 1 The Euclidean Algorithm gives , , , , and so we have d gcda, b 17 Bac-Substitution gives the sequence 1,, 3, 8, 19 and so we have a 19 b 8 d We have c/d 1/17 13, so we multiply the above equation by 13 to get a 47 b 104 c Thus one solution to the equation ax + by c is given by x, y u, v 47, 104 Since a/d 33/17 19 and b/d 765/17 45, the general solution is x, y u, v + b d, a d 47, , 19 where Z c Given positive integers a and b with gcda, b 1, find the largest integer c for which there do not exist positive integers x and y such that ax + by c Solution: We claim that the largest such integer is c ab We need to show that when c ab there do not exist positive integers x and y such that ax + by c, and when c > ab there do exist such positive integers x and y Suppose first that c ab Then one solution to ax + by c is given by x 0, y 0 b, 0 so, by the Chinese Remainder Theorem, the general solution is x, y b, 0 + b, a b b, a where Z To get x a > 0 we need > 0 since a > 0 but to get y b b > 0 we need b < b hence < 0 since b > 0, and so there are no positive integers x and y such that ax + by c Now suppose that c > ab Since gcda, b 1 we can choose s, t Z so that as + bt 1 and then we can let u sc and v tc to get au + bv c Then the general solution to the equation ax + by c is given by x, y u, v + b, a with Z Note that when x, y u, v + b, a we have x > 0 u b > 0 b < u < u b and we have y > 0 v + a > 0 a < v < v a Choose u b 1 Then < u b so that x > 0 and u b 1 au ab 1 c av ab 1 > ab av ab 1 v b so that y > 0

2 : a Let n 7 14 Find the prime factorizations of each of the numbers n, τn, σn and ρn Solution: Using the formula e p m! m p + m p + m p + we have 3 n ! 14! 13! Using the formula τ i i + 1 we have τn Using the formula σ i 1 + pi i we have σn Using the formula ρn n τn/ and noting that τn/ , we have ρn b Show that for n, m Z +, if ρn ρm then n m Solution: Let n, m Z + If ρn ρm 1 then n m 1 Suppose that ρn ρm Note that since ρn n τn/, it follows that for any prime p we have e p ρn ep n τn/ In particular the prime factors of ρn are equal to the prime factors of n Similarly, the prime factors of ρm are equal to the prime factors of m Since ρn ρm it follows that n and m have the same prime factors Let n i and let m l i where p 1, p,, p r are the distinct prime factors of n and m and i, l i Z + For each index i we have i τn/ e pi ρn e pi ρm l i τm/ and hence i τm τn l i If we had τm < τn then we would have i τm τn l i < l i for all i, but this would imply that τn i + 1 < l i + 1 τm Similarly, we cannot have τm > τn so we must have τm τn, hence i τm τn, l i l i for all indices i and so n m c For n Z + and l R, let σ l n d n d l so that, for example, we have σ 0 n τn and σ 1 n σn Show that for n, m Z + and 0 l R, if σ l n σ l m and σ l n σ l m then n m Solution: When n 1 we have σ l n 1 for all l R Let n r i where p 1, p,, p r are the distinct prime factors of n and each i Z + Since the positive divisors of n are the numbers of the form d r with 0 i i we have and σ l n d l d n p 1 1l p 1l 1 p l p rl r 0 0 r r 0 p l r 1 + l pi + p l i + + p il i σ l r 1 + l pi + p l i + + P i il r 1 + l pi + p l i + + i l r l i σ ln n l r 0 r r p r rl il + i 1l + + l + 1 il i Thus for all n Z + and all l R we have σ l n n l σ l n Given n, m Z + with σ l n σ l m and σ l n σ l m, and given 0 l R we have n l σ ln σ l n σ lm σ l m ml and hence n m

3 3: a Show that for all, n Z +, we have n gcd, n n Solution: Let d gcd, n When > n we have n 0 and hence n d n Suppose that n Note that n n!! n! n n 1! 1! n! n n 1 1 Choose s, t Z so that s + nt d Then we have d n s + nt n s n + nt n which is a multiple of n n 1 1 s n n nt n n s n t n b Show that for all a, b, c, d,, l Z, if ad bc 1 then gcda + bl, c + dl gcd, l Solution: Suppose ad bc 1 Let u gcd, l and let v gcd a+bl, c+dl Since u and u l it follows that u x + ly for all x, y Z In particular, u a + bl and u c + dl and so u v Since v a + bl and v c + dl it follows that v a + blx + c + dly for all x, y Z In particular, v a + bld c + dlb, that is v, and v c + dla a + blc, that is v l Since v and v l we have v u c For all positive integers a, b and c, if c ab then c gcda, c gcdb, c Solution: Write a p 1 1 p p n n, b p 1 1 p p n m n and c p 1 m m 1 p p n n Note that ab p p + p n n+ n gcda, c p 1 min{ 1,m 1} p min{,m } p n min{ n,m n} gcdb, c p 1 min{ 1,m 1} p min{,m } p n min{ n,m n} gcda, c gcdb, c p 1 min{ 1,m 1}+min{ 1,m 1} p min{,m }+min{,m } p n min{ n,m n}+min{ n,m n} Suppose that c ab so we have m i i + i for all i Fix an index i We consider three cases Case 1 If m i i then we have m i min{ i, m i } min{ i, m i } + min{ i, m i } Case If m i i then we have m i min{ i, m i } min{ i, m i } + min{ i, m i } Case 3 If m i i and m i i then we have m i i + i min{ i, m i } + min{ i, m i } In all three cases we have m i min{ i, m i } + min{ i, m i } Thus c gcda, c gcdb, c as required d Show that for all, l, n Z + we have gcd lcm, l, n lcm gcd, n, gcdl, n Solution: Let p be any prime number Let r e p, s e p l and t e p n Then e p gcd lcm, l, n min e p lcm, l, t min maxr, s, t { } minr, t if r s max minr, t, mins, t mins, t if s r max e p gcd, n, ep gcdl, n e p lcm gcd, n, gcdl, n Since this holds for all primes p, we obtain the required formula gcd lcm, l, n lcm gcd, n, gcdl, n

4 4: a Show that for all a, b, c Z, if a + b + c 3 0 then a b c 0 Solution: Let a, b, c Z with a, b, c 0, 0, 0 and suppose, for a contradiction, that a + b + c 3 0 If we had a 0 then we would have b, c 0, 0 and b c 3 so that b 3c, but this is not possible since e b is odd but e 3c is even If we had b 0 then we would have a, c 0, 0 and a c 3 so that a 3c, but this is not possible since e 3 a is even but e 3 3c is odd Thus we must have a 0 and b 0 Since a + b c 3 we have a + ab + b 3c, hence ab 3c a b and hence 8a b 3c a b But this is not possible since e 8a b is odd but e 3c a b is even b Show that there exist infinitely many primes of the form p 6 1 with Z + Solution: By the Division Algorithm, all integers are of the form 6+r for some, r Z with 0 r 5, Since , all integers are of the form 6+s for some, s Z with 1 s 4 Every prime p 3 is not a multiple of or 3, so every prime p 3 is of the form p 6 1 or p 6+1 for some Z + Suppose, for a contradiction that there are finitely many primes of the form p 6 1 with Z +, say p 1, p,, p l is the complete list of all such primes Let n 6p 1 p p l 1 None of the primes, 3, p 1, p,, p l is a factor of n since when n is divided by any of these primes q the remainder is q 1, not 0 and so all of the prime factors of n are of the form p 6+1 But since 6+16l+1 36l+6+6l+1 66l++l+1 it follows that the product of numbers hence, by induction, the product of finitely many numbers of the form is also of the same form Since all of the prime factors of n is of the form it follows that n is a product of finitely many numbers of the form and so n itself is of the form But n is not of the form because n 6p 1 p p l 1 which is of the form 6 1 c Show that there exists n Z with 1 n 100 such that the number 1 n! Solution: We have 100 1! 50 l 1!l! 50 l l 1! 50 and so , 50!! 50 l 1! which is a square 1 d Find all a, b Z + such that a a+b b b a l ! is a square l 1! ! l 1! Solution: Let a, b Z + with a a+b b b a Since a a+b b b a, the numbers a and b must have the same prime factors and, if a m i and b m l i, where the are distinct primes and i, l i Z +, then we have a a+b m a+b i and b b a m l b a i, so we must have i a + b l i b a for all indices i Since a + b > b a and i a + b l i b a we have i < l i for all i Since i < l i for all i we have a b, say b ac with c Z + Since a a+b b b a and b ac we have a a+ac ac ac a aac ac ac a Divide both sides by a ac a to get a a c ac a, that is a a c c 1 a It follows that a c c 1, and hence also that b ac a c c c 1 c c c+1 Note that c c 1 is a square if and only if c 1 is even or c is a square if and only if c c+1 is a square, and so the all solutions are of the form a, b c c 1/, c c+1/, where c Z + and c is odd or c is an even square Conversely, when c Z + and either c is odd or c is an even square, if we let a c c 1/ and b c c+1/ then a, b Z + and b ac and b b a ac ac a a ac a c ac a a ac a c ac 1 a ac a c c 1/ a a ac a a a a ac+a a b+a a a+b

5 5: Let S { x, y, z R 3 x + y + z 1 } and let S S \ {0, 0, 1} Define F : S R as follows Given a point x, y, z S, let L be the line in R 3 through 0, 0, 1 and x, y, z, let u, v, 0 be the point at which L crosses the xy-plane, and let F x, y, z u, v Also, let G F 1 : R S a Find a formula for F and G Solution: The points on L are of the form u, v, w 0, 0, 1 + t x, y, z 0, 0, 1 tx, ty, 1 + tz 1, t R The point at which L crosses the xy-plane occurs when w 0, that is when t 1 u, v, w tx, ty, 1 + tz 1 x 1 z, y 1 z, 0 Thus x u, v F x, y, z 1 z, y 1 z 1 z, and it is the point To find G F 1, given u, v R, let M be the line in R 3 through 0, 0, 1 and u, v, 0 The Gu, v is the point on M which lies in x, y, z S The points on M are of the form x, y, z 0, 0, 1 + t u, v, 0 0, 0, 1 tu, tv, 1 t, t R The points where M intersects S satisfy 1 x + y + z tu + tv + 1 t t u + t v + t t + 1, that is u + v + 1t t 0, so they occur when t 0 and t When t 0 we obtain the point x, y, z tu, tv, 1 t 0, 0, 1 and when t we obtain the point u x, y, z Gu, v tu, tv, 1 t, v, u +v 1 b Explain why F and G give a biective correspondence F : S Q 3 Q and G F 1 : Q S Q 3 Solution: The maps F and G give a biective correspondence F : S R 3 and G F 1 : R S When x, y, z Q 3, the formula for F shows that u, v F x, y, z Q, and when u, v Q the formula for G shows that Gu, v Q 3 and so the restrictions of F and G give a biective correspondence F : S Q 3 Q and G F 1 : Q S Q 3 c Use the biective correspondence in Part b to find all a, b, c, w Z such that a + b + c w Solution: Let a, b, c, w Z with a + b + c w If w 0 then a b c 0 Suppose that w 0 If 0 r Z is a common divisor of a, b and c, then r is also a divisor of w and we have a + b + c r r r w r Suppose that gcda, b, c 1 Since a +b +c w we have a + b + c w w w 1 so a w, b w, c w S Q 3 If a w, b w, c w 0, 0, 1 then a b 0 and c w Otherwise we have a w, b w, c w S Q 3 and so a w, b w, c w Gu, v u for some u, v Q, v, u +v 1 Taing u, v m, m l, where, l Z, we have a w, b w, c w m lm +l +m, +l +m, +l m +l +m +l +m a, b, c w m, lm, +l m If s a prime and p s w then since a, b and c have no common divisors, either p a or p b or p c and in either case it follows that p s +l +m Thus we have w +l +m, say +l +m wd, and so d a, b, c, w m, lm, +l m, +l +m We must have d gcdm, lm, + l m because gcda, b, c 1 and so It follows that all solutions a, b, c, w Z to the equation a + b + c w are of the form a, b, c, w r m, lm, +l m, + l +m for some r,, l Z, m Z +, d where d gcdm, lm, + l m