n + d + q = 24 and.05n +.1d +.25q = 2 { n + d + q = 24 (3) n + 2d + 5q = 40 (2)

Transcription

1 MATH 16T Exam 1 : Part I (In-Class) Solutons 1. (0 pts) A pggy bank contans 4 cons, all of whch are nckels (5 ), dmes (10 ) or quarters (5 ). The pggy bank also contans a con of each denomnaton. The total value s two dollars. How many combnatons of cons are possble? Do not go over all possbltes : you wll not receve credt! Soluton : Let n (resp. d and q) be the number of nckels (resp. dmes and quarters) n the pggy bank. Then we have n + d + q = 4 and.05n +.1d +.5q = Multplyng the second equaton by 0, we have to fnd the number of solutons (n, d, q) N 3 0 of { n + d + q = 4 (1) n + d + 5q = 40 () Note that n, d and q are strctly postve snce the pggy bank contans a con of each denomnaton. We frst fnd all ntegral solutons. Replacng equaton () by ()-(1), we get { n + d + q = 4 (3) d + 4q = 16 (4) Snce (d, q) = (0, 4) s a partcular soluton of (4), we get that all ntegral solutons (d, q) of (4) are gven by { d = 0 + 4t where t Z q = 4 t Substtutng ths nto (3) and solvng for n, we fnd that all ntegral solutons (n, d, q) are gven by n = 0 3t d = 4t where t Z q = 4 t Snce we are nterested n solutons wth n, d, q > 0, we must have So 0 3t > 0 and 4t > 0 and 4 t > 0 t < and t > 0 and t < 4 Snce t s an nteger, we get that 1 t 3. So there are three values for t. Ths leads to three solutons (n, d, q) N 3 0. Three. (a) (15 pts) Fnd all the ratonal ponts on the hyperbola x y = 1. (b) (5 pts) Fnd all the ntegral ponts on the hyperbola x y = 1. Soluton : (a) We see that (1, 0) s an ntegral pont on the hyperbola. Hence all the ratonal ponts on the hyperbola can be found be ntersectng the hyperbola wth ether a vertcal lne or a lne wth ratonal slope through the pont (1, 0). The vertcal lne through (1, 0) has equaton x = 1. So we have to solve { x = 1 x y = 1 1

2 We easly get that (1, 0) s the only soluton. Let q Q. The lne through (1, 0) wth slope q has equaton y = q(x 1). So we have to solve { y = q(x 1) (1) x y = 1 () Substtutng (1) nto (), we get or x (q(x 1)) = 1 (1 q )x + q x q 1 = 0 If 1 q = 0 then we get that x = 0. So x = 1 and y = 0. If 1 q 0, then we have a quadratc equaton n x and we know that x = 1 s a soluton. Snce the product of the roots of ax + bx + c s c, we get that a x = q 1 1 q 1 = q q 1 Substtutng ths nto (1), we fnd that ( q ) y = q(x 1) = q q 1 1 = q q 1 {( q All the ratonal ponts on x y ) } = 1 are gven by {(1, 0)} q 1, q q : q Q and q ±1 1 (b) Suppose that x, y Z wth x y = 1. Then (x y)(x + y) = 1 Snce x y, x + y Z we get that x y = x + y = 1 or x y = x + y = 1 So (x, y) = ( 1, 0) or (x, y) = (1, 0). The only ntegral ponts on x y = 1 are ( 1, 0) and (1, 0). 3. (a) (5 pts) Wrte 101 and 109 as a sum of two squares. (b) (15 pts) Use (a) to wrte 44036(= ) as a sum of two squares. You wll not receve credt for just wrtng as a sum of two squares. You must show your method. Soluton : (a) We easly get that 101 = 10 and 109 =

3 (b) We used the multplcatve property of the norm n Z[] to fnd the followng (for all a, b, c, d Z): (a + b )(c + d ) = N(a + b)n(c + d) = N((a + b)(c + d)) = N((ac bd) + (ad + bc)) = (ac bd) + (ad + bc) So usng (a), we get Multplyng by 4 =, we get = (10 )( ) = ( ) + ( ) = = = = ( 97) + ( 40) = = Remark: There s another way of rewrtng (a +b )(c +d ) as a sum of two squares by consderng a +b = N(a+b) and c + d = N(c d). Ths would lead to =

4 MATH 16T Exam 1 : Part II (Take-Home) Solutons You must work on these exercses by your self! You can use any book, calculator, computer, artcles on the nternet. I am the only person you can ask questons about the exam. 4. Let (x, y, z) be a Pythagorean trple. (a) (5 pts) Prove that one of the components s dvsble by 3. Hnt : Consder modulo 3 : what are the squares modulo 3? (b) (5 pts) Prove that one of the components s dvsble by 5 (consder modulo 5). (c) (5 pts) Prove that one of the components s dvsble by 4. Proof : (a) Note that the only squares modulo 3 are {0, 1}. Suppose that nether x nor y s dvsble by 3. Then x y 1 mod 3. Hence z x + y 1 mod 3 a contradcton snce s not a square modulo 3. Hence ether x or y s dvsble by 3. (b) Note that the only squares modulo 5 are {0, 1, 4}. Suppose that nether x nor y s dvsble by 5. Then x, y {1, 4} mod 5. If x y 1 mod 5 then z x + y mod 5, a contradcton snce s not a square modulo 5. If x y 4 mod 5 then z x + y 8 3 mod 5, a contradcton snce 3 s not a square modulo 5. Hence (x, y ) {(1, 4), (4, 1)} mod 5. So z x + y mod 5. Hence 5 z. So 5 z. (c) We know that there exst d, m, n N 0 such that gcd(m, n) = 1, m > n, not both m and n are odd and (x, y, z) {d(m n, mn, m + n ), d(mn, m n, m + n )}. WLOG, (x, y, z) = d(mn, m n, m + n ). Then x = mnd. Snce ether m or n s even, we see that x s dvsble by (0 ponts) Fnd all natural numbers a and b wth a b = b 3a. Soluton : One easly checks that a = 1 b = 1. So we may assume that a, b > 1. Hence we can consder the prme factorzaton of a and b. We only use the prmes that show up n ether factorzaton : a = p α and b = where p 1 < p < < p n are prmes, α, β N and not both α and β are zero for = 1,,..., n. Substtutng ths nto a b = b 3a, we fnd p bα = p 3aβ Usng unque prme factorzaton, we get p β bα = 3aβ for = 1,,..., n ( ) Suppose frst that 3a b. From (*), we get that α β for = 1,,..., n. Hence b = p β dvdes p α = a 4

5 So a = kb for some k N 0. Snce a b = b 3a, we get that (bk) b = b 3kb Hence (bk) = b 3k and so b 3k = k If k = 1 then b = 1, a contradcton. So k. But then b 3k > k : no solutons. Suppose next that 3a < b. From (*), we get that α < β for = 1,,..., n. Hence a = p α dvdes p β = b So b = ka for some k N wth k (snce a < b). Snce a b = b 3a, we get that Hence and so If k = then a 1 = 3 ; so a = 8 and b = ka = 8 = 16. If k = 3 then a 3 = 3 3 ; so a = 3 and b = ka = 3 3 = 9. If k = 4 then a 5 = 4 3 : no solutons. If k = 5 then a 7 = 5 3 : no solutons. If k then a k 3 > k 3 : no solutons. a ka = (ka) 3a a k = (ka) 3 a k 3 = k 3 One easly checks that (8, 16) and (3, 9) are ndeed solutons. So we stll have to prove the followng: Let m. Then m 3k > k for k =, 3, 4,... and m k 3 > k 3 for k = 6, 7, 8,... We prove ths by nducton. For k =, we get that m 3k = m 4 4 = 16 > 4 = = k. Assume that m 3k > k for some k. Then m 3(k+1) = m 3 m 3k > 3 k > 4k = (k) > (k ) For k = 6 we get that m k 3 = m 9 9 = 51 > 16 = 6 3 = k 3. Assume that m k 3 > k 3 for some k 6. Note that for k > 0, we have that k 3 3k k 3, k 3 3k k 3 k 3 and k 3 1 k 1 So m (k+1) 3 = m m k 3 > k 3 = k 3 + k 3 + k 3 + k 3 k 3 + 3k 3 + 3k = (k ) 3 All couples (a, b) N 0 wth a b = b 3a are {(1, 1), (8, 16), (3, 9)}. 5

6 6. (0 ponts) Let p be a prme wth p 3 mod 4. Prove that there exsts exactly one Pythagorean trple such that one of the components s p. Note that we consder the trple (x, y, z) to be the same as (y, x, z). Proof : Let (x, y, z) be a Pythagorean trple such that p {x, y, z}. We know that there exst d, m, n N 0 such that gcd(m, n) = 1, m > n, not both m and n are odd and (x, y, z) {d(m n, mn, m +n ), d(mn, m n, m +n )}. WLOG, (x, y, z) = d(mn, m n, m + n ). We consder three cases: x = p, y = p and z = p. Suppose frst that x = p. Then p = mnd, a contradcton snce p s odd. Suppose next that y = p. Then p = d(m n ) = d(m n)(m + n). Snce p s a prme and m n < m + n, we must have that d = m n = 1 and m + n = p. Solvng for m and n, we get that m = p and n = p 1 Substtutng ths nto the formulas for x, y and z, we fnd that x = mn = p p 1 ( ) ( ) = p 1 p 1 p, y = m n = p and z = m + n = + = p Note that ths s actually a prmtve Pythagorean trple. Suppose fnally that z = p. Then p = d(m + n ). Snce p s a prme and m + n > 1, we must have that d = 1 and m + n = p. So p s a sum of two squares, a contradcton snce p 3 mod 4. So we proved that there s exactly one Pythagorean trple such that p s one of the components, namely ( p 1 p ), p, 7. Let a, b, c N 0 wth gcd(a, b) = 1 and c ab. Prove that there s at most one couple (x, y) N 0 wth ax + by = c. Proof : Suppose there are at least two dfferent non-zero natural solutons to ax + by = c. Then the dstance between the X and Y -ntercepts of the lne ax + by = c must be strctly bgger than the dstance between those two solutons (strctly bgger because the ntercepts are not ponts n N 0). We ve seen n class (Proposton 1.3) that the dstance between the ntercepts s have c ab a + b whle the dstance between two consecutve solutons s a + b. Hence we c a + b ab > a + b So c > 1 or c > ab, a contradcton. ab Hence there s at most one couple (x, y) N 0 wth ax + by = c. 8. A trangular number s a number of the form n(n ) for some n N. (a) Prove that there are nfntely many trangular numbers that are perfect squares. (b) Fnd a trangular number that s a perfect square and s bgger than 1, 000, 000. Show your work! Proof : (a) Let t N. We easly get Note that t s a trangular number that s a perfect square t = m = n(n ) = m n + n = m n(n ) 4n + 4n = 8m (n ) 8m = 1 for some m, n N 6

7 So f t s a trangular number that s a perfect square then ( 1 + 8t, t) s a natural soluton of X 8Y = 1. Conversely, let (x, y) be a natural soluton of X 8Y = 1. Then x s odd snce x = 1 + 8y s odd. So x 1 N and ( ) ( ) x 1 x 1 t := y = x 1 = 1 ( ) ( ) x 1 x = 8 s a trangular number that s a perfect square. We have seen n class that the Pell-Fermat equaton X 8Y = 1 has nfntely many natural solutons. Hence there are nfntely many trangular numbers that are perfect squares. (b) We easly get that (3, 1) s the fundamental soluton of X 8Y = 1. By Theorem 1.33(b), all the solutons of X 8Y = 1 are gven by {( (3 + 8) k + (3 8) k, (3 + 8) k (3 ) } 8) k : k = 0, 1,,... 8 Recall from (a) that y s a trangular number that s a perfect square f x 8y = 1. So we are lookng for k N wth (3 + 8) k (3 8) k > = We easly get that k 5 does the job. So ( t := y (3 + 8) 5 (3 ) 8) 5 = = = 1189 = s a trangular number that s a perfect square. 7