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Slide 1: Solutions to the exercices of Elementary Number Theory by Edmund Landau Bruno Joyal January 8, 2008 Chapter 1 Exercise 1: Suppose b > 0 and b a. Put r0 = a, r1 = b, and determine r2 , r3 , ..., rn by the relations: r0 = r1 q1 + r2 0 < r2 < r1 r1 = r2 q2 + r3 0 < r3 < r2 ... rn−2 = rn−1 qn−1 + rn 0 < rn < rn−1 rn−1 = rn qn , rn+1 = 0 Show that rn , the last non-zero remainder, is the greatest common divisor of a and b. This process for finding the greatest common divisor is known as the Euclidean Algorithm. Proof: The Euclidean algorith is based on the fact that if r0 = r1 q1 + r2 , then (r0 , r1 ) = (r1 , r2 ). This is easily seen, as (r0 , r1 ) must divide both sides, and thus r2 ; since (r0 , r1 ) divides both r1 and r2 , (r0 , r1 ) is a common divisor of r1 and r2 ; and they can have no common divisor larger than (r0 , r1 ), as such a divisor would obviously divide r0 as well, contradicting the fact that (r0 , r1 ) is the greatest divisor of r0 and r1 . Thus we may reduce the problem of finding (a, b) to the problem of finding (r1 , r2 ), and so on, until the process terminates and we have found (a, b). Note that the algorithm works also if b | a, for then (a, b) is simply r1 . 1

Slide 2: Exercise 2: For arbitrary a and b, not both zero, consider all numbers of the form ax + by, where x and y take on all integer values. Let g be the smallest positive number in this collection. Show that g = (a, b). Proof: Numbers of the form ax + by are obviously all divisible by (a, b). Consider the smallest positive such number, and write it as g = ax1 + by1 . This number divides all other numbers of the form ax + by; for otherwise we would have ax + by = kg + r, r = a(x − kx1 ) + b(y − ky1 ), g > r > 0, contradicting the fact that g is the smallest positive number of the form ax + by. Thus g|a, g|b, and since (a, b)|g, we have g = (a, b). Exercise 3: Prove that (ma, mb) = m(a, b), where m > 0 and a and b are not both zero. Proof: Let (ma, mb) be the smallest number of the form (ma)x + (mb)y. We have (ma, mb) = max + mby = m(ax + by); and if ax + by > (a, b), then there is a number of the form (ma)x + (mb)y which is smaller than (ma, mb), contradicting the hypothesis; and since ax + by < (a, b) is clearly impossible, we have (ma, mb) = m(a, b). Exercise 4: Using the last exercise, prove that if a divides bc, and (a, b) = 1, then a divides c. Proof: Since we have ax + by = 1, we have acx + bcy = c; and since a | acx and a | bcy, a | c. Exercise 5: Show that (a + b, a − b) is either 1 or 2 if (a, b) = 1. Proof: (x, y) divides x + y and x − y; thus (a + b, a − b) divides 2a and 2b. Therefore, if (a, b) = 1, (a + b, a − b) can only be 1 or 2. n Exercise 6: Let Fn be the nth Fermat number, Fn = 22 + 1, for n = 0, 1, 2, .... Show that, if k > 0, Fn divides Fn+k − 2. Infer that any two Fermat numbers are relatively prime. Proof: Let n be given. For k = 1, the theorem is true, since Fn+1 − 2 = Fn (Fn − 2). Let us thus suppose the theorem true for k − 1. Then, for k, we n+k n+k−1 n+k−1 have Fn+k − 2 = 22 − 1 = (22 + 1)(22 − 1), which is divisible by Fn+k−1 − 2, which in turn, by the induction hypothesis, is divisible by Fn . Therefore, Fn divides Fn+k − 2 for all k > 0. 2

Slide 3: This result implies that any two distinct Fermat numbers Fn and Fn+k are relatively prime: since if d divides Fn , d > 2 (since Fn is odd), and it must divide Fn+k − 2 as well (which is divisible by Fn ). But if Fn+k − 2 is divisible by d, then Fn+k is not divisible by d; therefore Fn and Fn+k share no divisors other than 1. Note also that since there are infinitely many Fermat numbers, and they are all relatively prime to one another, there must be infinitely many prime numbers; for if there were only q primes, any collection of q + 1 Fermat numbers would include at least one pair of non-coprime numbers, by the so-called pigeonhole principle. Exercise 6: Prove that if n is odd, then n(n2 − 1) is divisible by 24. Proof: n(n2 −1) = (n−1)n(n+1), which is the product of three consecutive integers. Two of them are even, one of which must be divisible by 4; one of them is divisible by 3. Therefore their product is divisible by 24. Exercise 7: Show that in the so-called Fibonacci series, 1, 2, 3, 5, 8, ..., in which every term is the sum of the two preceding terms, two consecutive terms are always relatively prime. Proof: Let fn denote the nth Fibonacci number.The theorem is obvious for n = 1. Let it be true for n − 1. Let (fn , fn−1 ) = fn x + fn−1 y. Then (fn , fn−1 ) = (fn−1 + fn−2 )x + fn−1 y = fn−1 (x + y) + fn−2 x. But this is the smallest number of such form, for if there was a number fn−1 a + fn−2 b smaller than fn−1 (x + y) + fn−2 x, we would have: fn−1 a + fn−2 b = fn−1 a + (fn − fn−1 )b = fn b − fn−1 (a − b) < fn−1 (x + y) + fn−2 x = fn x + fn−1 y, contradicting the hypothesis that fn x + fn−1 y is the smallest number of such form. Thus (fn , fn−1 ) = (fn−1 , fn−2 ), and, by induction, (fn , fn−1 ) = 1 for all n > 0. 1 1 1 1 Exercise 8: Show that the sum Hn = 1 + 2 + 3 + 4 + ... + n is not an integer if n > 1. Proof: This sum may be written as m m m m+ 2 + 3 + ... + n , m with m the smallest common multiple of the numbers 1 through n. Since m contains, in its prime decomposition, exactly [log2 (n)] powers of two, m is even k 3

Slide 4: whenever k does not contain [log2 (n)] powers of two, and odd otherwise; so (n − 1) of the numbers m are even, and the number m for which k contains k k [log2 (n)] powers of two is odd. Thus the numerator of Hn is odd, while its denominator is even. Chapter 2 Exercise 1: Prove that if a > 2 and n > 1, then an − 1 is composite. Proof: Let us consider the sum Sn−1 = an−1 + an−2 + ... + a + 1. From aSn−1 = Sn−1 + an − 1, we know that an − 1 = (a − 1)Sn−1 . Since a > 2 and n > 1, both factors are > 1, so an − 1 is divisible both by Sn−1 and by a − 1. Exercise 2: Prove that T (a), the number of positive divisors of a, is odd if and only if a is a perfect square. Proof: If k a= pj rj j=1 is the canonical decomposition of a, then k T (a) = (rj + 1) j=1 is the number of positive divisors of a (since for every prime pj appearing in the decomposition of a, there are rj + 1 possibilities of chosing a power of pj ). Since we know that a number is a perfect square if and only if all numbers rj are even, we know that T (a) is odd if a is a perfect square, and even otherwise. We may generalize this result to show that a perfect nth power never has a number of positive divisors which is a mutiple of n (unless, of course, n = 1!), or to show that if a number has a prime number of positive divisors, then it is a power of a prime. Exercise 2: Suppose n > 0. a) Show that T (2n − 1) ≥ T (n). Proof: The theorem is obvious if n is prime, so let n be composite. For any divisor d of n, setting a = n , we have : 2n − 1 = (2a )d − 1 = (2a − 1)(2a(d−1) + d 2a(d−2) + ... + 2a + 1). Therefore, to every divisor d of n, we may map the unique divisor 2d − 1 (or, equivqalently, 2a − 1) of 2n − 1. This result implies that 2n − 1 can be prime only if n is prime. 4

Slide 5: b) Show that T (2n + 1) ≥ T ∗ (n), where T ∗ (n) is the number of odd divisors of n. 2n −1 Proof: Since we have 2n + 1 = 2 n −1 , we know that 2n + 1 will have at 2 least one divisor for every divisor of 2 − 1 which is not a divisor of 2n − 1; and 2n this is at least the number of divisors of 2n which are not divisors of n, which is T (2n) − T (n) = T ∗ (n). Exercise 4: Use Ex. 6 of Chap. 1 to give another proof of Theorem 18 (”there are infinitely many primes”). Proof: Ex. 6 of Chap. 1. Exercise 5: Suppose k to be a given integer greater than 2. a) If q1 − 1, q2 − 1, ..., qv − 1 are divisible by k, show that q1 q2 ...qv − 1 is divisible by k. Proof: Let q1 −1 = kc1 , q2 −1 = kc2 , ..., qv −1 = kcv . Then in the expansion of the product (kc1 + 1)(kc2 + 1)...(kcv + 1), every summand is divisible by k except the last one which is 1, so that q1 q2 ...qv − 1 is divisible by k. Alternate proof, by induction on v: The theorem is tautologous for v = 1. Let the theorem be true of any collection of v − 1 numbers qj − 1 which are divisible by k. Then, if q1 − 1, q2 − 1, ..., qv −1 is a collection of v numbers divisible by k, the product q1 (q2 q3 ...qv )−1 = (q1 − 1)(q2 q3 ...qv ) + (q2 q3 ...qv − 1) is definitely divisible by k, since (q1 − 1) is divisible by k by definition and q2 q3 ...qv − 1 is divisible by k, by hypothesis. b) If n > 0, show that there is a prime p such that k does not divide p − 1 and p divides (nk − 1). Proof: Let n be given and k > 2, since 1 and 2 divide p − 1 for every p > 2. Set nk − 1 = p1 p2 ...pc , where the primes are not necessarily distinct. If, for every prime p1 , p2 , ..., pc , k divides p − 1, we know by Ex. 5.a that k divides p1 p2 ...pc − 1, so that k must divide nk − 2, and so k must be 1 or 2, a contradiction. There is therefore a prime p which divides nk − 1 for which k does not divide p − 1. 5

Slide 6: c) By the method of proof of Theorem 18 (”any list of primes must be incomplete, since their product + 1 is not divisible by any prime in the list”), show that there are infinitely many primes p such that k does not divide p − 1. Proof: Let k > 2, otherwise the theorem is false. Let us assume that there are only m such primes : p1 , p2 , ...,pm , and consider Q = k(p1 p2 ...pm ) − 1 = pa1 pa2 ...pau , where the pa ’s are not necessarily distinct. Obviously Q is not divisible by k or by any of the primes p1 , ..., pm ; and if it is only divisible by primes for which k | p − 1, then k must divide pa1 pa2 ...pau − 1 as well, and so k must divide Q − 1 as well as Q + 1, and so k = 1 or 2, a contradiction. Therefore Q is divisible by a prime p, different from p1 , p2 , ..., pm , for which k p − 1; there are therefore infinitely many such primes. Exercise 6: If f (n) is a non-constant polynomial in n with integral coefficients, then f (n) is composite for infinitely many values of n. Proof: Let f (n) = cm xm + cm−1 xm−1 + ... + c0 . Whenever n is a multiple of c0 , f (n) is divisible by c0 , since every summand of f (n) is then divisible by c0 . Exercise 7: Prove that if x and y are real, [x + y] ≥ [x] + [y]. Proof: Let f (x) = x − [x], f (y) = y − [y]. We have 0 ≤ f (x) < 1 (for otherwise x − [x] > 1 implies x > [x] + 1, contradicting the definition x ≤ [x] < x + 1) Therefore, [x + y] = [[x] + [y] + f (x) + f (y)] = [x] + [y] + [f (x) + f (y)] ≥ [x] + [y]. We may generalise this result, using induction, to any number of real quan- tities, using the fact that [x + y] ≥ [[x] + y]. Note that if n is the number of quantities in the brackets, then the maximum amount by which the left side may exceed the right side is n − 1. Exercise 8: Use Theorem 27 and Ex. 7 to prove that if m > 0 and n > 0, then (m + n)! is divisible by m!n!. Infer that the product of any n consecutive integers is always divisible by n!. Proof: Let p be any prime which divides m!n!. By Theorem 27, it divides m!n! exactly ∞ m n + j j=1 pj p times. But it also divides (m + n)! ∞ m+n j=1 pj 6

Slide 7: times, and, by the last exercise we have, for every j: m+n m n ≥ j + j pj p p which proves the first statement of the exercise. For the second statement, we simply notice that since m!n! | (m + n)!, we have m! | (m+n!) . n! Exercise 9: If 1 ≤ r ≤ pn , pk | r, and pk+1 r, show that pn ! r!(pn − r)! is divisible by pn−k , but not by pn−k+1 . Proof: First, notice that this number is an integer, which we proved in the previous exercise. To prove the theorem, let us first set r = cpk , with p c and 0 < c ≤ pn−k . Let us denote by dr the power of p which appears in the canonical decomposition of r!, and similarily for dpn −r . Then we have, by Theorem 27 ∞ n−k cpk c dr = = c(pk−1 + pk−2 + ... + 1) + j=1 pj j=1 pj And also ∞ n−k pn − cpk pn−k − c dpn −r = = (pn−1 − cpk−1 ) + ... + (pn−k − c) + j=1 pj j=1 pj And for this last sum, since pj c, we have n−k n−k pn−k − c pn−k c = − j −1 j=1 pj j=1 pj p So that n−k c dpn −r = (pn−1 + ... + 1) − c(pk−1 + ... + 1) − − (n − k) j=1 pj Giving us dr + dpn −r = pn−1 + ... + 1 − (n − k). Since obviously, by Theorem 27, dpn = pn−1 + ... + 1, the theorem follows. 7

Slide 8: Chapter 3 Exercise 1: Suppose r ≥ 2, a1 > 0, ..., ar > 0, and let v be the least com- v v mon multiple of a1 , ..., ar . Show that if we write a1 = a1 , ..., ar = ar , then (a1 , ..., ar ) = 1. v Proof: By contradiction : suppose some prime p divides aj = aj for all v 1 ≤ j ≤ r; then p divides also ak , where ak is a number which contains the prime p with the highest power of a1 , a2 , ..., ar , in which case v would also be p common multiple of a1 , a2 , ..., ar , contradicting the hypothesis. Exercise 2: Suppose r ≥ 2, and a1 , ..., ar are not all zero, say a1 not zero. Let d1 = a1 and dn = (an , dn−1 ) for 2 ≤ n ≤ r. Show that dr = (a1 , a2 , ..., ar ). Proof: We prove that (a, (b1 , b2 , ..., br )) = (a, b1 , b2 , ..., br ); the rest follows from induction, since the theorem is tautologous for n = 2. We use a set- theoretic argument : if d divides the left side, it must divide both a and all the b’s, in which case it also divides the right side. If d divides the right side, by the same argument it must divide the left; therefore the equality stands, since both sides have exactly the same divisors. Exercise 3: If r ≥ 2 and a1 > 0, ..., ar > 0, let us denote (temporarily) the least positive common multiple of a1 , ..., ar by {a1, ..., ar}. Show that if a > 0, b > 0, and c > 0, then a) (a, {b, c}) = {(a, b), (a, c)}; b) {a, (b, c)} = ({a, b}, {a, c}); c) ({a, b}, {a, c}, {b, c}) = {(a, b), (a, c), (b, c)}. Proof: These identities are direct consequences of the properties of the basic set-theoretic operations of union and intersection. Let us map to every integer n the set P (n) of the greatest prime-powers which divide n, pk ∈ P (n) iff pk | n and pk+1 n. This map is bijective in virtue of the fundamental theorem of arith- metic. Noting that P ({a, b}) = P (a) P (b), and that P ((a, b)) = P (a) P (b), we know that we may now manipulate the numbers a, b, c exactly as if they were sets, and the operations of ”greatest common divisor” and ”least common multiple” exactly as if they were the operations of intersection and union on these sets. This gives us : a) (a, {b, c}) → P (a) P ({b, c}) = P (a) P (b) P (c) = P (a) P (b) P (a) P (b) → {(a, b), (a, c)}; 8

Slide 9: b) {a, (b, c)} → P (a) P ((b, c)) = P (a) P (b) P (a) = P (a) P (b) P (a) P (c) → ({a, b}, {a, c}); c) ({a, b}, {a, c}, {b, c}) → P (a) P (b) P (a) P (c) P (b) P (c) = P (a) P (b) P (a) P (c) P (b) P (c) → {(a, b), (a, c), (b, c)}. Chapter 4 For the exercises of this chapter, I will use two lemmas. Lemma 0.1 Let D(a, b) be 1 if a | b, 0 otherwise. Then we have b b−1 D(a, b) = − a a Proof The theorem is obvious if a = 1, so let a > 1. Then, if a | b, we have: b b−1 b b 1 − = − − =1 a a a a a And if a b, we have b = ka + r, 1 ≤ r < a, so that b b−1 ka + r ka + r − 1 − = − =k−k =0 a a a a Lemma 0.2 Let f (x) be any number-theoretic function. Then x x x f (j) = f (d). j=1 j j=1 d|j Proof The theorem is true for x = 1. But when it is true for x − 1, then, in the inductive jump to x, the left-hand side increases by x x−1 x x x−1 f (j) − f (j) = D(j, x)f (j) = f (d), j=1 j j=1 j j=1 d|x and obviously the right side does as well. Alternate proof A given f (d) will be in the sum on the right as many times as d divides numbers up to n, i.e., as many times as there are multiples of d up to x. Since x d is the number of multiples of d up to x, the theorem follows. 9

Slide 10: Exercise 1: If a > 0 and b > 1, show that S(a) S(ab) S(a)S(b) < ≤ . a ab ab Proof: Since S(ab) > bS(a) (all numbers in the sum b(1 + d2 + ... + dn ), where the d’s are the divisors of a, are distinct, since all the d’s are distinct; and they are all divisors of ab; the inequality comes from the fact that 1, for example, is also a divisor of ab but is not in the sum, since b > 1). Therefore S(ab) S(a) ab > a . From this it follows that a multiple of a perfect number cannot be a perfect number. The second part of the theorem follows from the fact that S(a)S(b) contains all the divisors of a and b, but might contain some of them too many times, since S(a)S(b) = (1 + da2 + ... + dan )(1 + db2 + ... + dbn ), so that if da = db for some d, it will be counted twice (once as da ∗ 1, once as 1 ∗ db ). Exercise 2: Show that an odd integer divisible by no more than two primes cannot be a perfect number. Proof: For such an odd integer n = p1 p2 , p1 > 2, p2 > 2, we have S(n) = p1 p2 +p1 +p2 +1. But p1 +p2 +1 < p1 p2 holds obviously for all numbers > 2, for otherwise if p1 + p2 + 1 ≥ p1 p2 we would have p1 + p2 + p11p2 ≥ 1; but 1 1 since 7 = 1 + 1 + 1 ≥ p1 + p2 + p11p2 , this is impossible. Therefore, if n = p1 p2 , 9 3 3 9 1 1 then S(n) < 2n. Exercise 3: Prove that the sum of the reciprocals of the positive divisors of a perfect number is equal to 2. a Proof: Let a be a perfect number. Since d runs through the positive divisors of a when d does, 1 d 1 = = d = 2. d a a d|a d|a d|a 10

Slide 11: Exercise 6: The number-theoretic function λ(a) (Liouville’s function) is defined thus : λ(a) = 1 if a = 1 or if a is the product of an even number of primes (not necessarily distinct), while λ(a) = −1 if a is the product of an odd number of primes (not necessarily distinct). Prove the following : a) If a > 0 and b > 0, then λ(ab) = λ(a)λ(b). Proof: Letting Ω(a) be the number of primes (not necessarily distinct) which compose a, we have λ(a) = (−1)Ω(a) for a > 0; thus, since Ω(ab) = Ω(a)+ Ω(b), if a > 0, b > 0, then λ(ab) = (−1)Ω(ab) = (−1)Ω(a) (−1)Ω(b) = λ(a)λ(b). b) λ(d) = 1 if a is a perfect square, 0 otherwise. d|a Proof: In words, the theorem means : a non-square integer has as many divisors with an odd number of primes than divisors with an even number of primes; squares have one more of the latter than of the former. First, we see that if a = p2k , then λ(d) = λ(1) + λ(p) + λ(p2 ) + ... + λ(p2k ) = 1, d|a since the summands alternate between 1 and -1 but the sum begins and ends with 1. We see also that if a = p2k+1 , then λ(d) = λ(d) + λ(p2k+1 ) = 1 − 1 = 0. d|a d|p2k Thus the theorem is true for all prime-powers; and, because λ is completely m r multiplicative (which we proved five minutes ago), if a is composite = pj j , j=1 then m λ(d) = λ(d). d|a j=1 d|prj j Therefore, if all r’s are even, a is a perfect square and the sum is 1, and if not all r’s are even, a is not a perfect square and the sum is zero. [x] x 1 c) For x ≥ 1, we have λ(n) = [x 2 ]. n=1 n Proof: If we can prove that it is true for all x in N, then it is true for all 1 1 x in R, since [x 2 ] = [x] 2 and n = [x] . Thus let us use induction on x. x n The theorem is true for x = 1; and when it is true for x − 1, then, during the jump to x, the right-hand side increases by 1 iff x is a perfect square, and by 0 11

Slide 12: otherwise, while the left-hand side increases by: x x−1 x x x x−1 λ(n) − λ(n) = D(n, x)λ(n) = λ(n), n=1 n n=1 n n=1 d|x which, by the previous exercise, is 1 iff x is a perfect square, and 0 otherwise. x λ(n) d) For x ≥ 1 we have <2 n=1 n Proof: Let us define the functions λ− (n) and λ+ (n) to mean, respectively, the number of numbers 1 ≤ k ≤ n for which λ(k) = −1 and for which λ(k) = 1. We have: x x x 1 x x x [x 2 ] = λ(n) = − n=1 n n n λ(n)=1 λ(n)=−1 x x x 1 x x λ(n) [x 2 ] ≤ − −1 =x + λ− (x) n n n=1 n λ(n)=1 λ(n)=−1 And also x x x 1 x x λ(n) [x 2 ] ≥ −1 − =x − λ+ (x) n n n=1 n λ(n)=1 λ(n)=−1 Since we have λ+ (x) ≤ x and λ− (x) ≤ x, we have : 1 x 1 [x 2 ] λ(n) [x 2 ] −1≤ ≤ + 1, x n=1 n x which completes the proof. x λ(n) We might as well go further and try to see if has a limit as x → ∞, n=1 n or if it oscillates between finite limits. We can, since y is fully multiplicative, use the Euler product for the harmonic series. This gives us : ∞ λ(ns ) λ(ps ) λ(p2s ) 1 = λ(1) + + + ... = (−1)s n=1 ns p ps p2s p 1− ps For s = 1, this necessarily converges in the interval [0, 1[. Two interesting identities which can easily be obtained from this representation are: ∞ λ(n2k ) = ζ(2k) n=1 n2k And ∞ λ(n2k+1 ) ζ(4k + 2) = . n=1 n2k+1 ζ(2k + 1) 12