MATH 145 Algebra, Solutions to Assignment 4

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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 161 975 1 + 86, 975 86 3 + 117, 86 117 + 5, 117 5 + 13, 5 13 4 + 0 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 765 33 + 119, 33 119 + 85, 119 85 1 + 34, 85 34 + 17, 34 17 + 0 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, 104 + 45, 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

: 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 7 14 7! 14! 13! 3 3 5 17 19 3 Using the formula τ i i + 1 we have 13+6+3+1 3 9+3+1 5 5+1 7 3 11 13 17 19 3 7+3+1 3 4+1 5 7 11 13 6+3+1 3 4+1 5 7 11 13 τn 3 4 3 5 3 Using the formula σ i 1 + pi + + + i we have σn 1 + + 1 + 3 + 3 + 3 3 1 + 5 + 5 1 + 171 + 191 + 3 7 40 31 18 0 4 9 3 3 5 7 31 Using the formula ρn n τn/ and noting that τn/ 4 3 16 9 144, we have ρn 3 3 5 17 19 3 144 88 3 43 5 88 17 144 19 144 3 144 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 0 1 1 0 1 1 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: 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 1 1 + nt n n s n 1 1 + 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 1 1+ 1 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: 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 6+5 6+1 1, 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 6 + 1 is also of the same form Since all of the prime factors of n is of the form 6 + 1 it follows that n is a product of finitely many numbers of the form 6 + 1 and so n itself is of the form 6 + 1 But n is not of the form 6 + 1 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 1 100 50, 50!! 50 l 1! which is a square 1 d Find all a, b Z + such that a a+b b b a l 50 100 1! is a square l 1! 50 50 50! 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: 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