Number Theory Proof Portfolio

Number Theory Proof Portfolio Jordan Rock May 12, 2015 This portfolio is a collection of Number Theory proofs and problems done by Jordan Rock in the Spring of 2014. The problems are organized first by general problems, individually assigned problems, and then problems I choose myself. Each section is also in numerical order by the text that they come from. The first four problems are the generally assigned problems to the class. They are in numerical order. 1 General Exercise 1 (5.4 A number L is called a common multiple of m and n if both m and n divide L. The smallest such L is called the least common multiple of m and n and is denoted by LCM(m, n For example, LCM(3,7)=21 and LCM(12,66)=132. (c) Give an argument proving that the relationship you found is correct for all m and n. Proof. First, let s start with the gcd. Let the gcd(a, b) = k for a, b, k N. By definition, k a and k b therefore k ab. Let s assume then that ab = l for l N. This means that l is k a multiple of a and l is a multiple of b. Therefore l is a common multiple of a and b. Now suppose a and b have another common multiple, let s call this m for m N. This means that m = ab where j is another divisor of a and b and j N. Notice that ab = l where k j k is the greatest common divisor, therefore k>l. This implies that m>l, making l the least common multiple. Thus lcm(a, b) = ab. gcd(a,b) 1

For the following exercise, we will be looking at modulo arithmetic, and what it means for a number to be congruent to a statement. We will be looking at how to solve these congruences for a variable. Exercise 2 (8.3 Find all congruent solutions to each of the following congruences. (a) 7x 3 mod 15 Notice that the inverse of 7 in mod 15 is 13. Therefore If we multiply the entire congruence, we will see that 13(7x 3 mod 15) 91x 39 mod 15 x 39 mod 15. Since the value of x for this congruence is 39, we can reduce this for modulo 15, therefore 39 9 mod 15. (c) x 2 1 mod 8 The values 1,3,5,and 7 are all solutions for this congruence because when these values are squared and divided by 8, they will have a remainder of 1. (e) x 2 3 mod 7 There are no solutions for this x because there is no integer that when squared has a remainder of 3 when divided by 7. For the following exercise, we will be introduced to the phi function. We will prove how to solve the phi function, and then do an example. Exercise 3 (11.3 Suppose that p 1, p 2,..., p r are the distinct primes that divide m. Show that the following formula for φ(m) is correct. Use this formula to compute φ(1000000 φ(m) = m(1 1 p 1 2

Proof. Base cases: Let m = p k 1 1. Then φ(m) = p k 1 1 p k 1 1 1 = p k 1 1 (1 1 p 1 Now let m = p k 1 2, then φ(m) = (p k 1 1 p k 1 1 1 )(p k 2 2 p k 2 1 2 ) = p k 1 2 (1 1 p 1 ) = m(1 1 p 1 Induction Hypothesis: If m 1 = p k 1 r, then φ(m 1 ) = m 1 (1 1 p 1 Now let m 2 = p k 1 r p k r+1 r+1. Therefore, φ(m 2 ) = φ(p k 1 1 )φ(p k 2 2 )... φ(p kr r )φ(p k r+1 r+1 We can then solve the phi functions, therefore φ(m 2 ) = (p k 1 1 p k 1 1 1 )(p k 2 2 p k 2 1 2 )... (p kr r p kr 1 r )(p k r+1 r+1 p k r+1 1 r+1 Here we will factor out all of the p i s up until p r such that φ(m 2 ) = (p k 1 r )(1 1 p 1 )(p k r+1 r+1 p k r+1 1 r+1 Finally, we can factor out a p k r+1 r+1 such that φ(m 2 ) = (p k 1 r p k r+1 r+1 )(1 1 p 1 )(1 1 p r+1 Notice that m 2 = p k 1 r p k r+1 r+1, so by substitution, φ(m 2 ) = (m 2 )(1 1 p 1 )(1 1 p 2 )... (1 1 p r )(1 1 p r+1 Thus we can conclude that φ(m) = m(1 1 p 1 To find φ(1000000), we need to know the prime factors of 1000000, which are 5 and 2. Therefore, φ(1000000) = (1000000)(1 1 5 )(1 1 2 ) = 1000000( 4 5 )( 1 2 ) = 1000000( 4 10 ) = 400000. In this next exercise, we are still working with modular arithmetic, and solving congruences. Exercise 4 (21.1 Determine whether each of the following congruences has a solution (all moduli are prime (a) x 2 1 mod 5987 Using Euler s Criterion, we see that, 1 (5987 1)/2 1 2993 = 1. Therefore, the legendre symbol ( 1 ) = 1. We know that -1 means this is a non residue mod 5987, so there are 5987 no solutions. (b) x 2 6780 mod 6781 First, we notice that 6780 1 mod 6781 because if we were to add 1 to both sides of the congruence, we would see that 6781 0 mod 6781. Here we can use Euler s Criterion to see that 1 (6781 1)/2 1 3390 = 1. Seeing that this equals 1, we know that 6780 is a quadratic residue in mod 6781, and therefore there is a solution. 3

The following section contains three problems that were given to be individually. They are also in numerical order. 2 Individual Assignments This first exercise works with M-world, which is a different number system than the one we are used. We will explore what it means to be prime in this different world, and how to find primes. Exercise 5 (7.6 Welcome to M-world, where the only numbers that exist are positive integers that leave a remainder of 1 when divided by 4. In other words, the only M- numbers are {1,5,9,13,17,21... }. In the M-world, we cannot add numbers, but we can multiply them, since if a and b both leave a remainder of 1 when divided by 4, so do their product. We day that m M-divides n if n = mk for some M-number k. And we say that n is an M-prime if its only M-divisors are 1 and itself. (a) Find the first six M-primes. The first six M-primes are 5,9,13,17,21, and 29. (b) Find a M-number n that has two different factorizations as a product of M-primes. The M-number 405 can be factored in these two ways, 405 = 5 81 = 9 45. Similar to the first section, in this exercise we will be working with more modular arithmetic, but this time we are finding how many (if any) solutions exist for each congruence. Exercise 6 (8.6 Determine the number of incongruent solutions for each of the following congruences. (a) 72x 47 mod 200 First we are going to find the gcd(72, 200) and if that number divides 47, then this congruence has that many solutions. If it does not divide 47, then there are no solutions. So 4

first, we use the Euclidean Algorithm to find the greatest common divisor like so, 200 = 72 2 + 54 72 = 54 1 + 18 54 = 18 3 + 0. Notice that since 18 is the last remainder before 0, that it is the greatest common divisor of 200 and 72. Unfortunately, 18 47, so this congruence has no solutions. (b) 4183x 5781 mod 15087 First, we need the gcd(15087, 4183) and we will find this using the Euclidean Algorithm. We can see that 15087 = 8143 3 + 2538 4183 = 2538 1 + 1695 2538 = 1695 1 + 893 1695 = 893 1 + 752 893 = 752 1 + 191 752 = 191 5 + 47 191 = 47 3 + 0. Notice that 47 is the last nonzero remainder, and therefore gcd(15087, 4183) = 47. Since 47 5781, we know that this congruence has 47 solutions. (c) 1537x 2863 mod 6731 First, we need to find the gcd(6731, 1537) using the Euclidean Algorithm. We can see that 6731 = 1537 4 + 583 1537 = 583 2 + 371 583 = 371 1 + 212 371 = 212 1 + 159 212 = 159 1 + 53 159 = 53 3 + 0. Notice that the gcd(6731, 1537) = 53. Since 53 2863, we know that this congruence has no solutions. 5

When working with congruences, to solve for quadratics, we cannot simply take the square root of both sides and call it a day. Here we will be working with roots of a congruence. Exercise 7 (17.3 In this chapter we described how to compute a k t h root of b mod m, but you may well have asked yourself if b can have more than one k t h root. It Can! If a is a square root of b mod m, then a is also a square root of b mod m. (a) Let b, k and m be integers that satisfy gcd(b, m) = 1 and gcd(k, φ(m)) = 1. Show that b has exactly one k t h root mod m. Proof. Let b, k, m Z such that gcd(b, m) = 1 and gcd(k, φ(m)) = 1. By way of contradiction, suppose x, y Z are unique solutions for the congruence x k b mod m and y k b mod m. We know that gcd(k, φ(m)) = 1 so by Bezout s lemma, there exist u, v Z such that ku + φ(m)v = 1. By Euler s Formula, we know that for integer a, a φ(m) 1 mod m. We can rewrite the equation as ku = 1 φ(m)v. thus, x ku x φ(m)v x mod m x mod m. Therefore, x ku b u mod m. By Euler s formula, we see that x φ(m)v+1 b u mod m, so x b u mod m. Without loss of generality, y b u mod m. Since gcd(b, m) = 1 and by the linear congruence theorem, there exists only one solution for x b u mod m and y b u mod m. Thus we have our contradiction and x y mod m. This last section contains three problems that I choose would be beneficial to this portfolio. I choose them because I believed they were a good representation of my learning for this course, and I felt that I had a good understanding of each of these exercises. 3 Free Choice The first exercise uses the greatest common divisor and Bezout s lemma to show that there exists solutions x, y Z for the equation ax + by = c when a and b are relatively prime to one another. Exercise 8 (6.5 Suppose that gcd(a, b) = 1. Prove that for every integer c, the equation ax + by = c has a solution in integers x and y. [Hint. Find a solution to au + bv = 1 and multiply by c.] Find a solution to 37x + 47y = 103. Try to make x and y as small as possible. 6

Proof. By the linear equation theorem, know there exist u, v Z such that au + bv = 1 for gcd(a, b) = 1. If we multiply both sides of this equation by a number c, we see that c(au + bv =1) auc + bvc =c We can now say that x = uc and y = vc such that ax + by = c. For example, let s consider 37x + 47y = 103. First we will find the gcd(37, 47) with the euclidean algorithm. Notice that 47 = 1 37 + 10 37 = 3 10 + 7 10 = 1 7 + 3 7 = 2 3 + 1 3 = 3 1 + 0 Notice that the gcd(37, 47) = 1. If we take each step of previous process and solve for the remainder, we can determine values for x and y in the equation 37x + 47y = 1. Let a = 47 and b = 37 such that 10 = a b 7 = b 3(a b) 7 = 4b 3a 3 = (a b) (4b 3a) 3 = 4a 5b 1 = ( 3a + 4b) 2(4a 5b) 1 = 11a + 14b. Notice here that the coefficients of a and b will be v and u respectively. Therefore vc = ( 11)(103) = 1, 133 and uc = (14)(103) = 1, 442 and hence for this solution, x = 1442 and y = 1133. The second exercise is a lengthy proof of the Chinese Remainder Theorem for 3 congruences. 7

Exercise 9 (11.9 In this exercise you will prove a version of the Chinese Remainder Theorem for 3 congruences. Let m 1, m 2, m 3 be positive integers such that each pair is relatively prime; gcd(m 1, m 2 ) = 1 gcd(m 2, m 3 ) = 1 gcd(m 1, m 3 ) = 1. Let a 1, a 2, a 3 be any 3 integers. Show that there is exactly one integer k in the interval 0 x < m 1, m 2, m 3 that solves these: x a 1 mod m 1 x a 2 mod m 2 x a 3 mod m 3. Can you figure out how to generalize this problem to deal with lots of congruences x a 1 mod m 1, x a 2 mod m 2... x a r mod m r? In particular, what conditions do the moduli m 1, m 2,..., m r need to satisfy? Proof. Let x a 1 mod m 1, x a 2 mod m 2, and x a 3 mod m 3 for m 1, m 2, m 3 +Z and for a 1, a 2, a 3 Z. Now let M 1 = m 2 m 3, M 2 = m 1 m 3, and M 3 = m 1 m 2. It is given that m 1, m 2, m 3 are relatively prime to each other, therefore gcd(m 1, m 1 ) = 1, gcd(m 2, m 2 ) = 1, and gcd(m 3, m 3 ) = 1. By Bezout s Lemma, we can say that there exist m 1 x 1 + M 1 y 1 = 1, m 2 x 2 + M 2 y 2 = 1, and m 3 x 3 + M 3 y 3 = 1 for y 1, y 2, y 3 Z. If we convert each of these equations back into modulo m form, we see that M 1 y 1 1 mod m 1, M 2 y 2 1 mod m 2, and M 3 y 3 1 mod m 3. By the linear congruence theorem, there is exactly one solution for y 1, and for y 2, and for y 3 such that 0 y 1 < m 1, 0 y 2 < m 2, and 0 y 3 < m 3. If we multiply both sides of each congruence by their respective a 1, a 2, a 3, we see that a 1 M 1 y 1 1 mod m 1, a 2 M 2 y 2 1 mod m 2, and a 3 M 3 y 3 1 mod m 3. If we add a 1 M 1 y 1 + a 2 M 2 y 2 + a 3 M 3 y 3 we see that a 2 M 2 y 2 + a 3 M 3 y 3 0 mod m 1, a 1 M 1 y 1 + a 3 M 3 y 3 0 mod m 2, and a 1 M 1 y 1 + a 2 M 2 y 2 0 mod m 3. Notice that this is because m 1 M 2 M 3, m 2 M 1 M 3, and m 3 M 1 M 3. Therefore x = a 1 M 1 y 1 + a 2 M 2 y 2 + a 3 M 3 y 3. Based off of this proof, the generalized form for any number of congruences is as follows x a 1 mod m 1, x a 2 mod m 2,..., x a r mod m r x = a 1 M 1 y 1 + a 2 M 2 y 2 + + a r M r y r. 8

Finally, the last exercise of this portfolio is another proof. This is on the last two cases of Quadratic Reciprocity. I think it is very important to know all of the cases, so I knew I wanted to include this proof in my portfolio. Exercise 10 (21.4 Finish the proof of Quadratic Reciprocity (part II) for the other two cases: primes congruent to 1 mod 8 and primes congruent to 5 mod 8. Proof. First we will start with P 1 mod 8. This means p = 8k + 1 for some k Z. The even integers between 1 and p 1 are 2, 4,..., p 1. The lower half of these evens are 2, 4,..., 4k. The upper half of these evens are 4k + 2, 4k + 4,..., 8k. To find the sun in the upper half, we will take half of the highest value, 8k, and subtract the lower half highest value, 4k. Therefore, 1 ((8k) (4k)) = 2k. This is the value of the negative signs, 2 so if we use the formula 2 (p 1)/2 1 2k mod p we see that 2 (p 1)/2 1 mod p. Therefore ( 2 ) = 1. Thus 2 is a QR if any prime is congruent to 1 mod 8. p Now let s work with p 5 mod 8. This means p = 8k + 5 for some k Z. All of the evens between 1 and p 1 are 2, 4,..., 8k + 4. The lower half is 2, 4,..., 4k + 2 and the upper half is 4k + 4, 4k + 6,..., 8k + 4. If we take half of the difference of the two highest values, 1 ((8k +4) (4k +2)), we see that the number of evens in the upper region is 2k +1. 2 This is also the number negative signs. Therefore 2 (p 1)/2 1 2k+1 mod p 1 mod p. By Euler s Criterion, 2 is a NR for all primes congruent to 5 mod 8. 9