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1 Chapter 4 The Arithmetic of Z In this chapter, we start by introducing the concept of congruences; these are used in our proof (going back to Gauss 1 ) that every integer has a unique prime factorization. We will also discuss the Euclidean Algorithm, a basic tool in computational number theory; we shall see later that the same method also works for polynomial rings K[X] over fields K. 4.1 Divisibility Just as subtraction was not defined for all pairs of natural numbers (in N, we could have defined m n for m, n N with m n), division is not defined for all pairs of nonzero integers. The theory of divisibility studies this observation in more detail. We say that an integer b Z divides a Z (and write b a) if there exists an integer q Z such that a = bq. Actually this definition makes sense in general domains (a commutative ring with 1 and without zero divisors), and even in monoids. A monoid is a set M on which a multiplication is defined (a map M M M) such that 1. multiplication is commutative and associative; 2. M contains a neutral element (1 M); 3. M is cancellative: if xy = xz for x, y, z M, then x = y. Examples for monoids are the nonzero natural numbers, the nonzero integers, as well as e.g. the following sets: 1. M = {1, 3, 5, 7,...} = 2N + 1; 2. M = {1, 2, 4, 6,...} = 2N {1}; 3. M = {1, 5, 9, 13,...} = 4N Carl-Friedrich Gauss: 1777 (Braunschweig, Germany) 1855 (Göttingen, Germany) 30

2 4. M = {1, 2, 4, 8, 16,...}. 5. the nonzero elements in a domain R. The main properties of the divisibility relation follow directly from the definition: Proposition 4.1. For all elements a, b, c of a monoid, we have 1. 1 a and a a; 2. if a b and b c, then a c. The proofs are immediate. For showing the second claim, observe that we have b = aq and c = br for q, r M; but then c = br = a(qr), hence a c. For domains, we have in addition a result involving the additive structure: Proposition 4.2. Let a, b, c be elements in some domain R. If a b and a c, then a (b ± c). Proof. These are formal consequences of the definition: We have b = aq and c = ar for q, r R; then b ± c = a(q ± r) implies that a (b ± c). Elements dividing 1 are called units; the units in Z are 1 and +1. First of all, they are units because they divide 1. Now assume that r Z is a unit; then there exists an element s Z with rs = 1. Clearly r, s 0, hence r, s 1. If r > 1, then 0 < s < 1, but there are no integers strictly between 0 and 1. Proposition 4.3. The set M of units in some monoid forms a group. Proof. We have to check the axioms. First, 1 M shows the existence of a neutral element. If u is a unit, then by definition there is some v M such that uv = 1. But then v = u 1 is also a unit, so inverses exist. Finally, if u and v are units, then uu = vv = 1 for some u, v M, and then (uv)(u v ) = 1, hence uv is a unit. Note that commutativity and associativity are inherited from M: if these axioms hold for all elements in M, then they surely will hold for all elements in M. We now give two important definitions. Let M be a monoid; then a non-unit p M is called irreducible if it only has trivial factorizations, i.e. if p = ab for a, b M implies that a M or b M. prime if p ab for a, b M implies that p a or p b. Being prime is a stronger property than being irreducible: Proposition 4.4. Primes are irreducible. 31

3 Proof. Let p be prime. We want to show it s irreducible, so assume that p = ab; we have to prove that a or b is a unit. Now clearly p ab, and since p is prime, we have p a or p b. Assume without loss of generality that p a. Then a = pc for some c M, hence p = ab = pbc, and since M is cancellative we deduce that 1 = bc. Thus b is a unit, and this concludes the proof. It is not true at all that irreducibles are always prime. It is basically in order to have lots of examples that we have dealt with monoids here. Consider e.g. the monoid M = {1, 2, 4, 6,...}; here 2 is irreducible since clearly 2 = 1 2 = 2 1 are the only factorizations of 2. On the other hand, 2 is not prime: we have since 36 = 2 18, but 2 6 because 6 is not divisible by 2 in M. Now we claim Proposition 4.5. There are infinitely many primes in Z. Proof. We give a proof by contradiction. Assume that there are only finitely many primes, namly p 1 = 2,..., p r, and consider the integer N = p 1 p r + 1. Then N > 1, hence it is divisible by a prime p. This prime p is not in our list: if we had p = p i, then p N and p N 1 = p 1 p i p r, hence p divides 1 = N (N 1): contradiction, because p is a prime, hence can t be a unit by definition. This is a really nice proof; unfortunately, it is not completely correct. In order to see what is missing, consider the monoid M = {1, 5, 9, 13,...}. We can imitate the proof above as follows: assume that p 1 = 5,..., p r is the list of all primes in M (the number 5 is indeed prime, as is easily shown: if 5 ab for a, b M, then 5 ab in Z, hence we may assume that 5 a in Z. But this means a = 5c, and we see that c must have the form 4n + 1 just like 5 and a: thus 5 a in M). Then we form N = p 1 p r + 4 M; clearly N > 1, so it must be divisible by some prime p, and as above we see that p p j for 1 j r. What s wrong with this proof? Let us start with the list of primes consisting only of 5. Then N = = 9. And although N > 1, N does not contain any prime factor because it is irreducible, but not prime! In fact, we have = 9 49, but Thus what is missing in our proof is the following: Proposition 4.6. Every integer n > 1 has a factorization into irreducible elements. Proof. This is clear if N = 2; now do induction on N and assume the claim is true for all N < n. If n is irreducible, everything is fine; if not, then n = ab, and by induction assumption both a and b are products of irreducibles, hence so is n. The proof of Prop. 4.5 shows that there are infinitely many irreducibles in Z: this is because every N > 1 in Z (or in M = 4N + 1) is divisible by an irreducible element. 32

4 The claim that there exist infinitely many primes in Z will be complete once we have proved that irreducibles are prime in Z. We will do this using congruences, which we will discuss next. 4.2 Congruences Congruences are a very clever notation invented by Gauss (and published in 1801 in his Disquisitiones Arithmeticae ) to denote the residue of a number a upon division by a nonzero integer m. More precisely, he wrote a b mod m if m (a b). for elements a, b, m Z. Examples mod 7; 10 0 mod 5; mod 3. The rules for divisibility can now be transferred painlessly to congruences: first we observe Proposition 4.7. Congruence between integers is an equivalence relation. Proof. Recall that a relation is called an equivalence relation if it is reflexive, symmetric and transitive. In our case, we have to show that the relation has the following properties: reflexivity: a a mod m; symmetry: a b mod m implies b a mod m; transitivity: a b mod m and b c mod m imply a c mod m for a, b, c Z and m Z \ {0}. The proofs are straightforward. In fact, a a mod m means m (a a), and every integer m 0 divides 0. Similarly, a b mod m is equivalent to m (a b); but this implies m (b a), hence b a mod m. Finally, if a b mod m and b c mod m, then m (b a) and m (c b), hence m divides the sum c a = (c b) + (b a), and we find a c mod m as claimed. Since defines an equivalence relation, it makes sense to talk about equivalence classes. The equivalence class [a] (or [a] m if we want to express the dependence on the modulus m) of an integer a consists of all integers b Z such that b a mod m; in particular, every residue class contains infinitely 33

5 many integers. In the special case m = 3, for example, we have [0] = {..., 6, 3, 0, 3, 6,...}, [1] = {..., 5, 2, 1, 4, 7,...}, [2] = {..., 4, 1, 2, 5, 8,...}, [3] = {..., 3, 0, 3, 6, 9,...} = [0], etc. Note that [0] = [3] = [6] =... (in fact, [0] = [a] for any a [0]), and similarly [1] = [4] =.... In general, we have [a] = [a ] if and only if a a mod m, that is, if and only if m (a a ). In the case m = 3, there were exactly 3 different residue classes modulo 3, namely [0], [1], and [2] (or, say, [0], [1], and [ 1] since [ 1] = [2]). This holds in general: Lemma 4.8. For any integer m > 1, there are exactly m different residue classes modulo m, namely [0], [1], [2],..., [m 1]. Proof. We first show that these classes are pairwise distinct. To this end, assume that [a] = [b] for 0 a, b < m; this implies b [a], hence a b mod m or m (b a): but since b a < m, this can only happen if a = b. Next, there are no other residue classes: given any class [a], we write a = mq+r with 0 r < m (the division algorithm at work again), and then [a] = [r] is one of the classes listed above. The set {0, 1, 2,..., m 1} is often called a complete set of representatives modulo m for this reason. Sometimes we write r + mz instead of [r]. The one thing that makes congruences really useful is the fact that we can define a ring structure on the set of residue classes. This is fundamental, so let us do this in detail. The elements of our ring Z/mZ will be the residue classes [0], [1],..., [m 1] modulo m. We have to define an addition and a multiplication and then verify the ring axioms. Addition : Given two classes [a] and [b], we put [a] [b] = [a + b]. We have to check that this is well defined: assume that [a] = [a ] and [b] = [b ]; then we have to show that [a + b] = [a + b ]. But this is easy: we have a a mz, say a a = ma, and similarly b b = mb. But then (a + b) (a + b ) = m(a + B) mz, hence [a + b] = [a + b ]. The neutral element is the residue class [0] = mz, and the inverse element of [a] is [ a], or, if you prefer, [m a]. In fact, we have [a] [0] = [a + 0] = [a] and [a] [ a] = [a+( a)] = [0]. The law of associativity and the commutativity are inherited from the corresponding properties of integers: since e.g. (a + b) + c = a + (b + c), we have ([a] [b]) [c] = [a] ([b] [c]). Multiplication : of course we put [a] [b] = [ab]. The verification that this is well defined is left as an exercise. The neutral element is the class [1]. Distributive Law: Again, ([a] [b]) [c] = [a] [c] [b] [c] follows from the corresponding properties of integers. 34

6 Theorem 4.9. The residue classes [0], [1],..., [m 1] modulo m form a ring Z/mZ with respect to addition and multiplication. Now that we have introduced the rings that we will study for some time to come, we simplify the notation by writing + and instead of and. Moreover, we will drop our references to classes and deal only with the integers representing them; in order to make clear that we are dealing with residue classes, we write instead of = and add a mod m at the end. What this means in practice is that we identify Z/mZ with the set of integers {0, 1,..., m 1}. ISBN (International Standard Book Num- Applications: ber) From the 1970s onward books are assigned an ISBN consisting of four parts: the first block specifies the country (or rather the language of the country), the second block gives information about the publishing company, the third about the book within that company, and the last digit is a check digit that is computed as follows: multiply the digits of the ISBN by 1, 2, 3,..., 10, starting on the left; the check digit is the integers 10 for which the sum of these product is 0 mod 11. The check digit X stands for 10. Example: compute the check digit of the ISBN ?. We find ? ? mod 11, and since 10 1 mod 11, this gives 4? 0 mod 11, so? = 4, and the complete ISBN is It is easy to see that if you type in an ISBN and make a single error, then the check digit will catch it; thus the ISBN is an example of a 1-error detecting code. 4.3 Unique Factorization in Z Before we can prove that irreducibles are prime, we need Proposition If p is irreducible, then Z/pZ is a field. Let us do some work first. For any rational number x Q, define the ceiling x of x to be the smallest integer x; by definition we have x x < x + 1. Put x = p a for a > 0 and multiply through by a; then p a p a < p + a, or, after subtracting p, 0 a p a p < a. Proof of Prop We have to show that if [a] [0], i.e., if 0 < a < p, then there exists a residue class [b] such that [ab] = [1]. This is trivial if a = 1, so assume a > 1 and put r 1 = p/a ; then 0 ar 1 p < a. If we had ar 1 p = 0, then the fact that p is irreducible implies a = 1 or a = p, contradicting our assumption. Thus 0 < ar 1 p < a. 35

7 If a 1 = ar 1 p = 1, then b = r 1 is the inverse of a: in fact, reducing ar 1 p = 1 modulo p gives ar 1 1 mod p, i.e., [a][r 1 ] = [1]. If a 1 > 1, then put r 2 = p/a 1 and repeat the above argument. If a 1 r 2 p = 1, then [a][r 1 r 2 ] = [1], and b = r 1 r 2 is the desired inverse of a; if not repeat this step. Since a i decreases by at least 1 in each step, the process must eventually terminate: if a n = 1, then we have the following equations: 0 < a 1 = ar 1 p < a, 0 < a 2 = a 1 r 2 p < a 1,... 0 < a n 1 = a n 2 r n 1 p < a n 2 a n = a n 1 r n = 1. These equations give rise to the following congruences: a 1 ar 1 mod p, a 2 a 1 r 2 mod p,... a n 1 a n 2 r n 1 mod p, 1 = a n a n 1 r n mod p. Starting with the last and working our way up we get 1 = a n a n 1 r n equiva n 2 r n 1 r n... a 1 r 2 r 3 r n ar 1 r 2 r n mod p. This shows that b r 1 r 2 r n mod p is the multiplicative inverse of a mod p. Here s an example: let us compute the inverse of the residue class [5] in Z/13Z. We find a = 5, p = 13, r 1 = 13 5 = 3, hence a 1 = ar 1 p = 2. Repeating this step provides us with r 2 = 13 2 = 7 and a 2 = a 1 r 2 p = 1. Thus r 1 r mod 13, and [5][8] = [1]. Remark. Fields F have very nice properties; one of them is that linear equations ax = b with a, b F and a 0 always have a unique solution: in fact, since a 0, it has an inverse element a 1 F, and multiplying through by a 1 we get x = a 1 b. This does not work in general rings: the equation 2x = 1 does not have a solution in Z/4Z, and the linear equation 2x = 2 has two solutions, namely x = [1] and x = [3]. The main result on which unique factorization will be built is the following: Proposition Irreducibles in Z are prime. Proof. Assume that p is irreducible and that p ab. If p a and p b, then [a] and [b] are invertible modulo p; but then [ab] has an inverse (if [a][r] = [b][s] = [1], then [ab][rs] = [1]), and therefore p ab. This proves that p a or p b. 36

8 This means in particular that every integer in Z has a factorization into primes. We will prove in a minute that this factorization is unique; here we will give an example showing that this is not obvious: consider the monoid M = {1, 5, 9, 13,...} of positive integers of the form 4n + 1 (this example is actually due to Hilbert). It is clear that every integer in M has a factorization into irreducibles, but it is not unique: for example, we have = 9 77, and 9, 21, 33 and 77 are all irreducible in M The reason why unique factorization fails is the existence of irreducibles that aren t prime. The theorem of unique factorization asserts that every integer has a prime factorization, and that it is unique up to the order of the factors. Theorem Every integer n 2 has a prime factorization n = p 1 p r (with possibly repeated factors). This factorization is essentially unique, that is: if n = p 1 p r and n = q 1 q s are prime factorizations of an integer n, then r = s, and we can reorder the q j in such a way that p j = q j for 1 j r. A partial result in the direction of Theorem 4.12 can already be found in Euclid s elements; the first explicit statement and proof was given by Gauss in Proof. We already know that prime factorizations exist, so we only have to deal with uniqueness. This will be proved by induction on min{r, s}, i.e. on the minimal number of prime factors of n. We may assume without loss of generality that r s. If r = 0, then n = 1, and n = 1 = q 1 q s implies s = 0. Now assume that every integer that is a product of at most r 1 prime factors has a unique prime factorization, and consider n = p 1 p r = q 1 q s. Since p 1 is a prime that divides n = q 1 q s, it must divide one of the factors, say p 1 q 1 (after rearranging the q i if necessary). But q 1 is prime, so its only positive divisors are 1 and q 1 ; since p 1 is a prime, it is a nonunit, and we conclude that p 1 = q 1. Canceling p 1 shows that p 2 p r = q 2 q s, and by induction assumption we have r = s, and p j = q j after rearranging the q i if necessary. Remark. There is a simple reason for doing induction on the minimal number of prime factors and not simply on the number of prime factors of n: the fact that the number of prime factors of an integer is well defined is a consequence of the result we wanted to prove! In fact, in M = {1, 5, 9,...} numbers may have factorizations into irreducibles of different lengths: an example is 225 = = Some Applications We have already seen that integers are squares of rational numbers if and only if they are squares of integers. Here we shall use unique factorization to show that p is irrational. For assume not: then p = r 2 /s 2 for r, s N, and assume that r and s are coprime (if they are not, cancel). Thus ps 2 = r 2. Thus p r 2, and since p is prime, we must have p r, say r = pt. Then ps 2 = p 2 t 2, hence 37

9 s 2 = pt 2. But then p s 2, hence p s since p is prime, and this is a contradiction, since we now have shown that p r and p s although we have assumed that they are coprime. 4.4 Greatest Common Divisors in Z We will now introduce greatest common divisors: we say that d is a greatest common divisor of a, b Z and write d = gcd(a, b) if d satisfies the following two properties: 1. d a, d b; 2. if e Z satisfies e a and e b, then e d. We can use the unique factorization property to give a formula for the gcd of two integers. Before we do so, let us introduce some notation. We can write an a Z as a product of primes. In fact fact we can write a = ± p ai i, where the product is over all irreducible elements p 1, p 2, p 3,..., and where at most finitely many a i are nonzero. In order to avoid the ± in our formulas, let us restrict to positive integers from now on. Lemma For integers a, b N we have b a if and only if b i a i for all i, where a = p ai i and b = p bi i are the prime factorizations of a and b. Proof. We have b a if and only if there is a c N such that a = bc. Let c = p ci i be its prime factorization. Then c i 0 for all i, and a i = b i + c i, hence b a is equivalent to a i b i for all i. Here s our formula for gcd s: Theorem The gcd of two nonzero integers is given by a = p ai 1 and b = p bi i d = p min{ai,bi} i. Proof. We have to prove the two properties characterizing gcd s: 1. d a and d b. But this follows immediately from Lemma If d a and d b, then d d. In fact, write down the prime factorization d = p d i i d d. of d. Then d a and d b imply d i min(a i, b i ) = d i, hence Now assume that d and d are gcd s of a and b. Then d d by 2. since d is a gcd, and d d since d is a gcd, hence d = ±d. 38

10 For the ring Z of integers, we have much more than the mere existence of gcd s: the gcd of two integers a, b Z has a Bezout representation, 2 that is, if d = gcd(a, b), then there exist integers m, n Z such that d = am + bn. Theorem 4.15 (Bezout s Lemma). Assume that d = gcd(a, b) for a, b Z; then d has a Bezout representation. Proof. Consider the set D = az + bz = {am + bn : m, n Z}. Clearly D is a nonempty set, and if c D then we also have c D. In particular, D contains positive integers. Let d be the smallest positive integer in D; we claim that d = gcd(a, b). There are two things to show: Claim 1: d is a common divisor of a and b. By symmetry, it is sufficient to show that d a. Write a = rd + s with 0 s < d; from d = am + bn we get s = rd a = r(am + bn) a = a(rm 1) + b(rn), hence s D. The minimality of d implies s = 0, hence d m. Claim 2: if e is a common divisor of a and b, then e d. Assume that e a and e b. Since d = am + bn, we conclude that e d. The existence of the Bezout representation is a simple consequence of the fact that d D. Note that the key of the proof is the existence of a division with remainder. Bezout s Lemma can be used to give an important generalization of the property p ab = p a or p b of primes p: Proposition If m ab and gcd(m, b) = 1, then m a. Proof. Write ab = mn; by Bezout, there are x, y Z such that mx + by = 1. Multiplying through by a gives a = max + aby = max + mny = m(ax + ny), that is, m a. Finally, observe that canceling factors in congruences is dangerous: we have 2 8 mod 6, but not 1 4 mod 6. Here s what we re allowed to do: Proposition If ac bc mod m, then a b mod m gcd(m,c). Proof. We have m (ac bc) = c(a b). Write d = gcd(m, c), m = dm, c = dc, and note that gcd(m, c ) = 1. From dm dc (a b) we deduce immediately that m c (a b); since gcd(m, c ) = 1, we even have m (a b) by Prop 4.16, i.e. a b mod m gcd(m,c). 4.5 The Euclidean Algorithm In most modern textbooks, Unique Factorization is proved using the Euclidean algorithm; it has the advantage that a similar proof can also be used for other rings, e.g. polynomial rings K[X] over fields K. The Euclidean algorithm 2 Etienne Bezout: 1730 (Nemours, France) 1783 (Basses-Loges, France) 39

11 is a procedure that computes the gcd of integers without using their prime factorization (which may be difficult to obtain if the numbers involved are large). Moreover, it allows us to compute a Bezout representation of this gcd (note that our proof of Thm was an existence proof, giving no hint at how to compute such a representation). Given integers m and n, there are uniquely determined integers q 1 and r 1 such that m = q 1 n + r 1 and 0 r 1 < n. Repeating this process with n and r 1, we get n = r 1 q 2 + r 2 with 0 r 2 < r 1, etc. Since n > r 1 > r 2 >... 0, one of the r i, say r n+1, must eventually be 0: Example: m = 56, n = 35 m = q 1 n + r 1 (4.1) n = q 2 r 1 + r 2 (4.2) r 1 = q 3 r 2 + r 3 (4.3)... r n 2 = q n r n 1 + r n (4.4) r n 1 = q n+1 r n (4.5) 56 = = = = 2 7 Note that the last r i that does not vanish (namely r 3 = 7) is the gcd of m and n. This is no accident: we claim that r n = gcd(m, n) in general. For a proof, we have to verify two things: Claim 1: r n is a common divisor of m and n. Equation (4.5) shows r n r n 1 ; plugging this into (4.4) we find r n r n 2, and going back we eventually find r n r 1 from (4.3), r n n from (4.2) and finally r n m from (4.1). In particular, r n is a common divisor of m and n. Claim 2: if e is a common divisor of m and n, then e r n. This is proved by reversing the argument above: (4.1) shows that e r 1, (4.2) then gives e r 2, and finally we find e r n from (4.5) as claimed. The Euclidean algorithm does more than just compute the gcd: take our example m = 56 and n = 35; writing the third line as gcd(m, n) = 7 = and replacing the 14 by 14 = coming from the second line we get gcd(m, n) = 21 1 ( ) = Now 21 = gives gcd(m, n) = 2 ( ) 1 35 = , and we have found a Bezout representation of the gcd of 56 and 35. This works in complete generality: (4.4) says r n = r n 2 q n r n 1 ; the line before, which r n 1 = r n 3 q n 1 r n 2, allows us to express r n as a Z-linear combination of r n 2 and r n 3, and going back we eventually find an expression of r n as a Z-linear combination of a and b. 40

12 Bezout representations have an important practical application: they allow us to compute multiplicative inverses in Z/pZ. In fact, let [a] denote a nonzero residue class modulo p; since Z/pZ is a field, [a] must have a multiplicative inverse, that is, there must be a residue class [b] such that [ab] = [1]. Since there are only finitely many residue classes, this can always be done by trial and error (unless p is large): for example, let us find the multiplicative inverse of [2] in Z/5Z: multiplying [2] successively by [1], [2], [3], [4] we find [2] [3] = [6] = [1]; thus [2] 1 = [3] (we occasionally also write mod 5). Computing the inverse of [2] in Z/pZ is actually always easy: note that we want an integer b such that [2b] = [1]; but [1] = [p + 1], hence we can always take b = p+1 2. In general, however, computing inverses is done using Bezout representations. Assume that gcd(a, p) = 1 (otherwise there is no multiplicative inverse), compute integers x, y Z such that 1 = ax + py; reducing this equation modulo p gives 1 ax mod p, i.e., [a][x] = [1], or [a] 1 = [x]. 4.1 Prove that 2 n(n + 1) for all n N a) using induction b) directly. 4.2 Prove that 3 n(n 2 1) for all n N. Generalizations? 4.3 Prove that 8 (n 2 1) for all odd n N. 4.4 Prove or disprove: if n ab and n a, then n b. 4.5 Show that there are arbitrary long sequences of composite numbers (Hint: observe that and can be seen to be composite without performing any division; generalize!) 4.6 Show that divisibility defines a partial order on Z by writing a b if b a. 4.7 Show that, for integers a, b, c, d, m Z with m > 0, we have a b mod m = a b mod n for every n m; a b mod m and c d mod m = a+c b+d mod m and ac bd mod m; a b mod m = ac bc mod m for any c Z. 4.8 Show that there are infinitely many primes of the form 3n Try to extend the above proof to the case of primes of the form 3n + 1 (and 5n 1). What goes wrong? 4.10 Show that primes p = c 2 + 2d 2 satisfy p = 2 or p 1, 3 mod Show that primes p = c 2 2d 2 satisfy p = 2 or p 1, 7 mod Show that primes p = c 2 + 3d 2 satisfy p = 3 or p 1 mod 3. 41

13 4.13 Compute d = gcd(77, 105) and write d as a Z-linear combination of 77 and Check the addition and multiplication table for the ring Z/3Z: Compute addition and multiplication tables for the rings Z/5Z and Z/6Z Compute the multiplicative inverse of [17] in Z/101Z Compute gcd(2 m 1, 2 n 1) for small values of m, n 1 until you discover a general formula for d Let U 1 = U 2 = 1, and U n+1 = U n + U n 1 denote the Fibonacci numbers. Find a formula for gcd(u m, U n) Show that the Fermat numbers F n = 2 2n + 1 are pairwise coprime Show that there are infinitely many primes of the form p = 4n Show that there are infinitely many primes of the form p = 3n Compute gcd(x 2 + 2x + 2, x 2 x 2) over Z/mZ for m = 2, 3, 5 and 7, and find its Bezout representation Let a, b N be coprime, and let r N be a divisor of ab. Put u = gcd(a, r) and v = gcd(b, r), and show that r = uv Assume that M p = 2 p 1 is a prime. List the complete set of (positive) divisors of N p = 2 p 1 M p, and compute their sum. Conclude that if M p is prime, then N p is a perfect number (a number n is called perfect if the sum of its (positive) divisors equals 2n). Euler later proved that every even perfect number has the form 2 p 1 M p for some Mersenne prime M p. It is conjectured (but not known) that odd perfect numbers do not exist Compute the last two digits of For primes p {3, 5, 7, 11, 13}, compute A ( p 1 )! mod p. Can you find a 2 pattern? If not, compute B A 2 mod p. Formulate a conjecture Check which of the primes p {3, 5, 7, 11, 13} can be written as p = a 2 + b 2 with integers a, b N (e.g. 5 = ). Formulate a conjecture. 42

14 4.28 For some small primes p = 4n + 1, compute the smallest residue (in absolute value) of a mod p, where a = 2n n. (Example: for p = 5, we have n = 1 and 2 1 = 2 2 mod 5.) Compare with the results from the preceding Exercise. Formulate a conjecture and test it for a few more primes a) Given a 5-liter jar and a 3-liter jar and an unlimited supply of water, how do you measure out 4 liters exactly? b) Can you also measure out 1, 2 and 3 liters? c) Which quantities can you measure out if you are given a 6-liter and a 9-liter jar? d) Formulate a general conjecture. Can you prove it (at least partially)? 4.30 Show that a number n = d n... d 1d 0 = d n10 n d d 0 satisfies the congruence n d n d 1 + d 0 mod 9: the residue class modulo 9 of any integer is congruent to the sum of the digits of n Show that a number d n... d 1d 0 = d n10 n +...+d 1 10+d 0 satisfies the congruence n ( 1) n d n d 2 d 1 + d 0 mod Invent a simple method to compute the residue class of n = d n... d 1d 0 = d n10 n d d 0 modulo Compute the last digit of Compute the last two digits of

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