Number Theory Review for Exam #2

Transcription

1 Number Theory Review for Exam #2 1. Applications of Continued Fractions Be able to: Use convergents of a continued fraction to find the smallest solution to Pell s equation x 2 + dy 2 = 1 Find other solutions to Pell s equation for a given solution Use continued fractions to explain properties of musical scales Recall that a diophantine equation is an equation involving only integer coefficients and exponents, and variables that only take on integer values. A special type of diophantine equation that we studied are known as Pell equations. Any equation of the form x 2 dy 2 = n where both d is a positive integer, and n an integer is a Pell equation. If d is a perfect square (say d = e 2 ), then it is fairly simple to solve the equation x 2 dy 2 = n. Note that n = x 2 dy 2 = x 2 e 2 y 2 = (x ey)(x + ey). So there exist integers a and b such that n = ab, a = x ey and b = x + ey. Solving for x and y in terms of a, b,e we get x = (a + b)/2 and y = (b a)/2e. Thus, the solution to x 2 dy 2 = n are simply the x and y of the form x = (a + b)/2 and y = (b a)/2e where a, b are postive integers such that n = ab, and x and y are positive integers. If d is not a perfect square, then life is a bit different. We showed that if x and y are positive integers such that x 2 dy 2 = n, then x/y is a convergent of the continued fraction for d. So the way that we find solutions to x 2 dy 2 = n (when d is not a perfect square) is to find the continued fraction expansion of d, compute its convergents p k /q k, and see for which k is x = p k, and y = q k a solution. In the case that n = 1, then we can in fact find all of the solutions to x 2 dy 2 = 1. Here s how: 1. Find the convergents p k /q k of d 2. Find the smallest k such that x = p k, y = q k is a solution to x 2 dy 2 = For each positive integer l compute (p k + dq k ) l. This will simplify to a l + db l for some integers a l and b l. 4. Then one can show (here s how We know that p 2 k dq2 k = 1. Factoring gives: (p k dq k )(p k + dqk ) = 1. Raising to the lth power gives (p k dq k ) l (p k + dq k ) l = 1. Substitution gives: (a l db l )(a l + db l ) = 1. Multiplying the LHS out gives a 2 l = db2 l = 1. ) that x = a l, and y = b l is a solution to x 2 dy 2 = 1, and that these are all of the solutions. In class we discussed why standard music scale has 12 notes per octave. The basic idea was that people noted that the sound coming out of tubes of length x, 2x, and 3x all sounded good together. Thus, they could build a sequence of tubes of lengths between 1 and 2 by starting with a tube of length 1, tripling the length of a tube and then halving it until its length was between 1 and 2, and repeating the process, until they arrived at a tube that was very close to one of length 1. Be able to explain why this leads to approximating log 3 (2) be a rational e/f, the physical meanings of e and f, and the relationship to the continued fraction for log 3 (2). 1.1 Find all solutions to the Pell equation x 2 4y 2 = 65. Here: d = 4, and d is a perfect square. So we can factor to get (x 2y)(x + 2y) = 65. We list all ways to factor 65, and then solve for x and y. Factorization x 2y x + 2y x y

2 1.2 Below is a table of convergents for 13 k p k q k (a) Find a solution to x 2 13y 2 = 4 We try x = p k and y = q k for various values of k. k = 3 works. So x = 11 and y = 3 is a solution. (b) Using the convergents, find another solution to x 2 13y 2 = 4. We try x = p k and y = q k for other values k. k = 5 works. So x = 119 and y = 33 is a solution. 1.3 (a) Use the table in Problem 1.2 to find the least positive solution to x 2 13y 2 = 1. bf We try x = p k and y = q k for various values of k. k = 9 works. So x = 649 and y = 180 is a solution. (b) Say that your answer to (a) is x = a and y = b. Compute (a + 13b) 2 to find another solution to x 2 13y 2 = 1. We compute ( ) 2 = (180) = So x = and y = is a solution. 1.4 You are given a poster (that is 1 foot by 1 foot), and asked to produce a family of copies of the poster of various dimensions ranging from 1 by 1 to 2 by 2. The poster photocopier only allows you to magnify by 500% or reduce by 50%, assuming no two posters in the family are almost the same dimensions. (a) What dimensions of posters would you produce? I can multiply by 5 and divide by 2. So I would produce boards of dimension 1, 5/4 = 1.25, 25/16 = , 125/64 = , 625/512 = , 3125/2048 = , 15625/8192 = /2 14 = , etc. (b) How many posters would you have in your family? Why? 28 boards. Each time I construct a new board by multiply by 5 and dividing by a power of 2 until the dimension is between 1 and 2. I do this until I get a dimension that is close to 2. So I need to find integers e and f so that 5 e /2 f 1. Taking log 2 of both sides, I want to find integers e and f so that e log 2 (5) f 0. Or equivalently, log 2 (5) f/e. The convergents of log 2 (5) are 2/1, 7/3, 65/28. Taking e = 1 and f = 2 only gives 2 boards, taking e = 7 and f = 3 only gives 3 different board sizes. So to have a larger variety of board sizes I went up to the next convergent. (c) How many posters would you have to make to be able to produce your family of posters? I will need to make boards of dimension 5, 5 2, 5 3, 5 4, 5 5,..., 5 6, 5 28 In addition, I will need to half each of these of these 2, 4, 6, 9, 11, 13, etc., respectively. In general the board of dimension 5 k will have to be halved k log 2 (5) times. So I will need to make a total of k=1 k log 2(5). (d) What does this have to do with continued fractions? What number are you trying to approximate by a rational? We want to approximate log 2 5 by a rational f/e See answer to b for complete details

3 2. Fundamental properties of integers Be able to: State definitions, and main results. Work with, and prove results concerning the gcd. Compute the gcd of 2 numbers using the Euclidean algorithm. Prove Euclid s Lemma. Prove the infinitude of primes. A prime is a positive integer that has exactly 2 prime divisors. The primes less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. The Fundamental Theorem of Arithmetic asserts that every integer n 2 can be expressed as a product of primes. Moreover, up to reordering of the primes, the expression is unique. For example, the prime factorization of 19 is 19, and the prime factorization of 100 is or The greatest common divisor of the integers a and b (not both 0) is defined to be the largest integer d such that d a and d b. We write gcd(a, b) for the greatest common divisor of a and b. We say that the integers a and b are relatively prime, provided gcd(a, b) = 1. The two most important properties of the gcd are: If d a and d b, then d gcd(a, b). There exists integers u and v such that au + bv = gcd(a, b). Using the first property, one can show that if a = bq + r where q and r are integers, then gcd(a, b) = gcd(b, r). This leads to the Euclidean Algorithm for computing the gcd. Using the second property (here s how Suppose that a and b are relatively prime integers and that a (bc). Since a and b have gcd equal to 1, there exist integers u and v with au + bv = 1. Multiplying by c gives auv + bcv = c. Now a auv, and a bcv. So a c. ) one can prove the following: If a and b are relatively prime integers and c is a integer such that a bc, then a c. This implies, In particular, Euclid s Lemma: if p is a prime, and p bc, then either p b or p c. One can use Euclid s Lemma to show that there are an infinite number of primes, here s how Suppose to the contrary that there are only a finite number of primes. Then we can list them: as p 1, p 2,..., p k. Consider n = p 1 p 2 p k + 1. By the Fundamental Theorem of Arithmetic, n has a prime divisor p. So p = p i for some i. Now p n, and p p 1 p 2 p k. So by the 2 out of 3 rule, p divides 1. Hence p = ±1. But this is impossible since p is a prime. So we have reached a contradiction. Problems: 2.1 Find integers u and v so that 154u v = gcd(154, 1111). Answer: [ ] [ ] [ ] [ Thus u = 36 and v = 5 works. ] [ Explain why there do no exist integers u and v so that 123u + 456v = 1. Note that 3 divides 123 and 3 divides 456. So if such integers existed, then 3 would have to divide 1, which it clearly doesn t. 2.3 Show that if a and b are relatively prime integers, and c is an integer such that a c and b c, then (ab) c. Suppose that gcd(a, b) = 1, a c and b c. We know that there exist integers u and v with au + bv = 1. Multiplying by c gives auc + bvc = c. Now ab (auc) because b c, and ab bvc because a c. Thus, by the 2 out of 3 rule, ab divides c. ]

4 2.4 a and b are integers with a = 13b Show that gcd(a, b) gcd(b, 14). (Hint: it suffices to show that gcd(a, b) b and gcd(a, b) 14). Let d = gcda, b. By definition d a and d b. Since 14 = a 13b, the 2 out of 3 rule implies that d 14. Clearly, d b. Since d b and d 14, d divides the greatest common divisor or b and Explain why has a prime divisor other than 2, 3, 7, 11, 13. By the Fundamental Theorem of Arithmetic, has a prime divisor p. If p {2, 3, 7, 11, 13}, then p would divide 1 (a contradiction). Hence p / {2, 3, 7, 11, 13}. 3. Solving diophantine and modular equations Be able to: Determine all solutions to a linear diophantine equation. Solve linear mod equations. Use the Chinese remainder theorem. Use Fermat s Little theorem. Use Lucas Lemma. Let a, b and c be integers (with not both a and b equal to 0), and consider the (linear) diophantine equation ax + by = 0. (1) Let d = gcd(a, b). Then x = b/d and y = a/d is a solution to (1) because a(b/d)+b( a/d) = ab/d ab/d = 0. In fact for each integer k, x = kb/d and y = ka/d is a solution to (1) because a(kb/d) + b( ka/d) = 0. To see that these are all the solutions to (1), suppose that x and y are solutions. Then ax = by. Dividing by d we get (a/d)x = (b/d)y. Note gcd(a/d, b/d) = 1 (because d is the greatest common divisor of a and b). Thus, by Euclid s Lemma, (a/d) y. Say y = k(a/d). Then solving for x we see that x = (b/d)k. So x and y have the desired form. Now consider the (linear) diophantine equation ax + by = c (2) If gcd(a, b) doesn t divide c, then the system (??) doesn t have any (integer) solutions because gcd(a, b) divides the LHS but not the RHS of equation (??). Now suppose that gcd(a, b) c, say dk = c. By The Euclidean Algorithm, there exist integers u and v such that au + bv = d. Multiplying by k gives, a(uk) + b(vk) = dk = c. Thus, (??) has a solution, namely x = x p = ku and y = y p = kv. Note that if x = x and y = y are also solutions, then ax p + by p = c and ax + by = c. Subtracting these two equations, shows that a(x 0 x p ) + b(y 0 y p ) = 0. Hence, by (??), there is an integer l such that x p x 0 = bl/d and y p y 0 = al/d. In other words, all of the solutions to (??) are: x = x p + bl/d and y = y p al/d (l an integer). Now let n be a positive integer, and a and b integers. We say that a and b are congruent mod n provided n divides b a. We write this as a b (mod n). The modular equation ax b (mod n) has a solution if and only if n (ax b), or equivalently, if and only if there is an integer y such that ny = ax b. Thus the modular equation ax b (mod n) and the diophantine equation ax ny = b go hand in hand. This leads to the following result: ax b (mod n) has a solution if and only if gcd(a, n) b. If x p is a solution to ax b (mod n), then so is x p + kn/gcd(a, n) for each integer k.

5 If x p is a solution to ax b (mod n), then all the (incongruent) solutions are x p + kn/gcd(a, n) (k = 0, 1, 2,..., gcd(a, n) 1). Sometimes we need to work with different moduli. More specifically, sometimes we need to find one integer x that satisfies all the equations x a 1 (mod n 1 ) (3) x a 2 (mod n 2 ) (4). (5) x a t (mod n t ) (6) simultaneously. The most common case is when n 1, n 2,..., n t are pairwise relatively prime (i.e. gcd(n i, n j ) = for all i j). This case leads to the so-called Chinese Remainder Theorem where asserts that (in this case) there is a unique solution x with 0 x < n 1 n 2 n t, and all other solutions can be obtained by adding multiples of n 1 n 2 n t to this particular solution. In practice, we can solve (??) for t = 2, 3 by repeatedly solving a sequence of linear diophantine equations. For example, let s look at the case t = 3. The integers that solve x a 3 (mod n 3 ) all have the form x = a 3 + n 3 y for some integer y. Now we look at the 2nd equation. We want y so that a 3 + n 3 y (mod n 2 ) or equivalently n 3 y a 2 a 3 (mod n 2 ). This equation has a solution because gcd(n 2, n 3 ) = 1 We take y to be one solution. Then all integers that solve the 2nd and 3rd equation have the form x = a 3 + n 3 y + n 2 n 3 z where z is an arbitrary integer. We plug this into the first equation and get (after simplification) that n 2 n 3 z a 1 a 3 n 3 y (mod n 1 ). This has a solution because gcd(n 2 n 3, n 1 ) = 1. Taking z to be one sol n, we see that x = a 3 + n 3 y + n 2 n 3 z solves all 3 equations in (??). Also, we can add any multiple of n 1 n 2 n 3 to x and still have a solution. Thus, there will be a unique solution x with 0 x < n 1 n 2 n 3. A very important result is Fermat s Little Theorem. This asserts that if p is a prime, and a is an integer, then a p a (mod p). Another result is Lucas Lemma. Let p be a prime, and let a and b be integers with base-p representations a k a k 1 a 0 and b k b k 1 b 0, respectively. Then ( ) ( ) a ak b Problems: b k ( ak 1 b k 1 ) ( a0 3.1 Find all integer solutions to 36x + 21y = 6 [ ] [ ] [ b 0 ) ] [ (mod p) ] [ So we have that ( 5) = 1. Mutiplying by 6 gives 36(18)+21( 30) = 6. So a particular solution is x = 18 and y = 30. The general solution is x = k/gcd(36, 21) = k, and y = 30 36/gcd(36, 21) = 30 12k. 3.2 Find all integer solutions to 36x + 21y = 7 gcd(36, 21) = 3. Since 3 doesn t divide 7, there are no solutions. 3.3 The integers a and b have gcd 5, and x = 3, y = 6 is a solution to ax + by = 15. Find all (integer) solutions to ax + by = 15 x = 3 + kb/5 and y = 6 ka/5 3.4 Find all (incongruent) solutions to 144x 36 (mod 1200) [ ] [ ] [ So gcd(144, 1200) = 48. Since 48 does not divide there are no solutions. ] ]

6 3.5 Find all (incongruent) solutions to 144x 7 (mod 1200) Again, gcd of 144 and 1200 does not divide 7. So there are no solutions. 3.6 Let a and n be positive integers. Prove that ax b (mod n) has at least one solution for each integer b if and only if a and n are relatively prime. First suppose that a and n are relatively prime. Then gcd(a, n) = 1. Since 1 divides every b, ax b (mod n) has at least one solution for each b. Now suppose that for each b, ax b (mod n) has a solution. Then gcd(a, n) b for every possible integers b. This required gcd(a, n) to be 1. Hence a and n are relatively prime. 3.7 Show that if the equations (??) have a solution, then gcd(n i, n j ) (a i a j ) for each i and j. Suppose that we have a solution x. Then n i divides x a i and n j divides x a j. So the gcd of n i and n j divides both x a i and x a j. Thus gcd(n i, n j ) (x a j ) (x a i ) a i a j. 3.8 Find the smallest positive integer x such that x 1 (mod 25), x 2 (mod 36), and x 3 (mod 49). Show your work guess and check isn t a valid method. Start with x = y. Any such x will solve x 3 (mod 49). Next we find a y for which x solves the 2nd equation. Need y 2 (mod 36) or equivalently, 13y 1 (mod 36). (Using the Euclidean Algorithm on 13 and 36, we get that 13 ( 11) = 1. Sow y = 11 is a solution. Now any x of the form z = z solves the last two equation. We just need to find a z to solve the first equation. Or equivalentlym, we need to find z so that z 1 (mod 25). This simplifies down to 14z 9 (mod 25). Using the Euclidean Algorithm on 14 and 25 we get the 14(9) + 25( 6) = 1. So 14(9) 1 (mod 25). Thus z = 9 9 works. Since 81 6 (mod 25), it is simpler to take z = 6. Now x = = As 0 x < , we have our answer. If x wasn t in this range, we would add an appropriate multiple of to get it into this range. 3.9 Find another solution to the equations in Problem works It is known that 151 is a prime. Find, with justification, the remainder when is divided by 151. by Fermat s Little Theorem (mod 151). Now (2 150 ) 10 (1) 10 1 (mod 151). So the remainder with is divided by 151 is 1. Use Lucas Lemma to find the remainder when ( 80 26) is divided by 3. Let s write 80 and 26 in base 3. Now 80 = So the base-3 representation of 80 is Also 26 = So the base-3 representation of 26 is By Lucas Lemma ( ) ( )( )( )( ) (mod 3) So the remainder with ( 80 26) is divided by 3 is Factorization methods Be able to: Prove that if n has no prime divisor p with p n, then n is is prime. Explain when and why Fermat Factorization works. Use Fermat Factorization to find a factorization of n. Use Pollard rho to find a factorization of n. Explain why, in Pollard rho, if g ij > 1, then g ij is a non-trivial factor of n.

7 Given a positive integer n, one of the fundamental problems is to find its prime factorization. The first step in this is to determine whether or not n is prime. If n is prime, we are done. The prime factorization of n is n. If n is not prime, then we try to find factors a and b with 2 a, b, n 1 and n = ab. Then we find the prime factorization of a and b. We know of two tests to see if n is prime. The depends on the following result: If n is not prime, then n has a prime divisor p with p n. We can prove this as follows: Suppose that n is not prime. Then there exist integers p and q with n = pq, and 1 < p, q < n. Without loss of generality we may assume that p q. So p 2 pq = n. So p n. Any prime divisor of p will be a prime divisor of n and no larger than n. The contrapositive of the above statement is: If n has no prime divisor less than or equal to n, then n is prime. Another test depends on Fermat s Little Theorem which asserts that if p is a prime and a is an integer such that p a, then a p 1 1 (mod p). Thus, if we can find an integer a with 1 a < n, such that a n 1 1 (mod n), then n is not a prime. Suppose we know that n is not prime. How can we find a factor a of n (other than a = 1 or a = n? We discussed two different methods. The first is Fermat Factorization, and we only use this in the case n is odd. (If n is even, we have the factorization n = 2 (n/2)). We note that anytime we can write n as a difference of perfect squares, say n = x 2 y 2, then we get a factorization n = (x + y)(x y) of n. Conversely, anytime we have a factorization n = ab of n, then we can write n as a difference of squares, namely, n = ()x + y)/2) 2 + ((x y)/2) 2. Thus, finding a factorization of n is the same thing as writing n as the difference of two squares. Fermat factorization, runs through the integers x n, until one finds an x such that x 2 n is a perfect square, say y 2, with x y 1. Then n has the factorization n = (x y)(x + y). Note that Fermat factorization is efficient when n has two factors that are nearly equal. Another factorization method we know is the Pollard rho method. The idea here is to construct random integers x 1, x 2,..., x s with x i x j (mod n). Then one computes g ij = gcd(x i x j, n). This can be done quickly by using the Euclidean algorithm. If one gets lucky and some g ij > 1, then g ij n (becauseg ij is the greatest common divisor of x i x j and n) and 1 < g ij < n (because n doesn t divide x i x j ) Thus, this g ij will be a nontrivial factor of n. One can show that with high probability, if n has a prime divisor p and s p, then some g ij > 1. Thus, the Pollard rho method works (with high probability) in the case that n has a small prime divisor. Often, one constructs the x i s as follows. Choose an integer x 1 (often x 1 = 2). Then x j+1 is the remainder when x 2 j + 1 is divided by n. Problems: 4.1 Error in statement. The original statement have p n. That is is incorrect. As n = 35 would be a counter example. Let n be a positive integer with n 3 mod 4. Prove that if no prime p with p 3 (mod 4) is a divisor of n, then n is prime. Assume that there is no prime p with p 3 (mod 4) which divides n. Since n 3 (mod 4), n is odd. So 2 doesn t divide n. Let n = p 1 p 2 p k be the prime factorization of n. Each p i is odd, so is either 1 or 3 mod 4. If one p i is 3 mod 4, then we are done. Otherwise, each p i is 1 mod 4. But then n (mod 4), and we have a contradication. So it must be the case that at least one of the p i is 3 mod Use Problem 4.1 to show determine whether or not 163 is prime. Indicate which primes p you considered. 163 is not divisible by 2,3,5,7 So we need only chekc for primes divisors less than 163/7, that is, less than 21. Primes less than 21 that are 3 mod 4 are: 3, 7, 11, 19, 23. None of these is a divisor of 163. So 163 is prime. 4.3 It is known that (mod 1027). Explain (in complete detail), why this shows that 1027 is not prime. If 1027 is prime, then by Fermat s Little Theorem = 2 (mod 1027). But since (mod 1027), we see that = 4 2 (mod 1027) 4.4 Use Fermat Factorization to factor So we start with x = 81. With x = 81 we get that = 4. So = (81 2)(81 + 2) =

8 4.5 Use the Pollard rho method to find a nontrivial factor of x 1 = 1, x 2 = 5, x 3 = 26 x 4 = 677 x 5 = 2778 x 6 = 695 x 7 = 5244 x 8 = x 9 = 2389 x 10 = 7379 x 11 = 5251 x 12 = 6612 Note that g 12,9 = 41. So 41 is a nontrivial factor of In fact =