Solutions to Problem Set 4 - Fall 2008 Due Tuesday, Oct. 7 at 1:00

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1 Solutions to 8.78 Problem Set 4 - Fall 008 Due Tuesday, Oct. 7 at :00. (a Prove that for any arithmetic functions f, f(d = f ( n d. To show the relation, we only have to show this equality of sets: {d N } = { n d N }. So we have the following string of equivalences. x {d N } y N s.t. xy = n y N s.t. y n and n y = x x { n d N }. (b Prove that if g is another arithmetic function, then ( n f(d g = ( n f d d g(d. Here we can manipulate the index in the summation to show that the two sums are equal. Note that this method can also be used above. ( n f(dg = f(dg(e = f(eg(d = ( n f g(d. d d de=n de=n. The Riemann Zeta function is one of the most important functions in number theory (and the subject of a million dollar research prize!. It is defined for complex arguments s as ζ(s := n n s = p prime p s, although the above formulas only converge for R(s >. (a Prove that the sum and product formulas for ζ(s are actually equal. For this problem I will ignore some of the tricky convergence issues to give a more intuitive explanation. We start by rewriting the index of the sum using the fundamental theorem of arithmetic! ζ(s = n n s = n=p r p r p r k k (p r pr pr k k s = r i N {0} r i 0 (p r. i i s i

2 Note that the requirement that a sequence of nonnegative integers converge to zero is the same as assuming that only finitely many are nonzero. This shows that the last sum is in fact the same as the middle one. So now we do a massive distribution of this sum of products. r i N {0} r i 0 (p r = i i i s (p r = i r 0 i s i r 0 (p s i r. and now we recognize the geometric series and compute it; = i (p s i = p prime p s. (b Prove that the inverse of the zeta function can be written as ζ(s = n µ(n n s. We compute the product of ζ(s with this summation. ζ(s n µ(n n s = n = r mn=r n s n ( m s µ(n n s = r µ(n n s = m n ( r s µ(n. mn=r ( m s µ(n n s Now we notice that the interior sum is just the sum of µ(n over the divisors of r. This is shown in the book to equal zero unless r =, in which case it is one. So we have that ζ(s µ(n n s = s + s s 0 + =. n So we have shown that the product of the sum with ζ(s is identically one, which means that the sum must be ζ(s. 3. (Niven..7 Show that 6! 63! (mod 7. Note that we have that 70! (mod 7 from Wilson s Theorem. information to the problem at hand, 63!(64(65(66(67(68(69(70 = 70! (mod 7. We calculate this product mod 7: Relating this (64(65(66(67(68(69(70 ( 7( 6( 5( 4( 3( ( This immediately gives that 63! 70! (mod 7. Now, the same trick can be used to finish the problem. From the work above, we know 6!(6(63 6!( 9( 8 6!(7 6! (mod 7.

3 4. (Niven..5 Prove that (p! p (mod P, where P = p. Hint: Use the Chinese Remainder Theorem and Wilson s Theorem. We know the proof of the summation formula for P : P = p = (p p = p We can assume p is odd by computing the case for p =, which holds because! =. Next we compute (p! modulo the two factors. Since p (p!, it is clear that (p! 0 (mod p. We also have (p! (mod p by Wilson s Theorem. Now, note p 0 (mod p, p (mod p. Since p prime implies ( p, p =, we may apply Chinese Remainder Theorem to claim that because (p! and p are equivalent modulo the two relatively prime factors of P, they are congruent mod P. 5. The harmonic sums are defined as H n := m n (m,n= and we write H n = An B n as fractions. For example, H p = p for any prime p, and H = = It is a fact that if n >, H n is never an integer, and thus B n. (a Prove that p A p for any prime p. Hint: Pair the terms i and p i. We want to show that p A p for prime p. Since we are assuming the fraction representation is in lowest terms, this means that p A p and p B p. So first we note that if we collect all of the terms together by taking the common denominator of (p!, we have that p (p!. Now we see what the hint gives us. For any i p, i + (p i + i p = = p i i(p i i(p i (i (i + (... (p i (p i + (... (p p =. (p! That last fraction is just to show that when we combine all of these pairs together over the common denominator (p!, each term has a factor of p in the numerator. So the entire sum will have a factor of p in the numerator, and since there is no such factor in the denominator, it will not get canceled when reducing the fraction. m, p.

4 (b Prove that n A n for all n. Here we can use the exact same trick as above. The key difference is our common denominator should be chosen to be C n = m. m n (m,n= Then this denominator has the necessary quality that it is coprime to n, so just as above, we know that the factor of n we get in the numerator will not be canceled when reducing the fraction. (Bonus Prove that p A p for any prime p 5. (Bonus Find and prove a formula for A n mod n for all n. 6. (Niven.4.4 Show that the Carmichael number 56 is composite by showing that it is not a strong probable prime for base. For the strong probable prime test, we write 56 as the product So we need to calculate 35 (mod 56. We can calculate this by successive squaring and reduction: n n (mod So 35 = (mod 56. Now, if we successively square and reduce this number, we get So we have found that 80 (mod 56. Now, that alone not enough to show that 56 is Carmichael, but we not that in the line before we got the, we found that So this shows that we have found a square root of (mod 56 that is not or. Thus 56 cannot be prime. 7. Recall that a composite integer n is a Carmichael number if it is a probable prime for all bases, so a n a (mod n for all a. (a Suppose that n is squarefree. Prove that n is a Carmichael number if and only if (p (n for every prime divisor p n. Hint: Use the Chinese Remainder Theorem on the congruence a n a (mod n. We are given that n is squarefree, so we can write its prime factorization: n = p p... p l. Then n is a Carmichael number if and only if a Z, a n a (mod n.

5 Using CRT, we have that this is true if and only if, for each i l, a Z, a n a (mod p i. Since this equation is trivial for a 0, we can assume we can divide by a, and so it is equivalent to a Z such that p a, a n (mod p i. But we know that every unit mod p has order dividing p, and in fact, there is some element with order p. So choosing a as this number implies that (p i (n for each i. So we have shown the equivalence. (b Prove that every Carmichael number is squarefree. Hint: If n has a square factor, you just need to find one a such that a n a (mod n. So now consider some prime divisor of n with p n. Then for a = n p Z, a = n p = n n p 0 a (mod n. So n is not a Carmichael number. 8. (Niven.4.5 Show that 047 is a strong probable prime for. We use the same method in problem 6: 046 = 03, so now we calculate 03 by successive squaring. Since 04 is the tenth power of two, I ll just find 04 and divide by. n n (mod So 03 = 0 (mod 047, so 047 is a strong probable prime for the base. However, it should be noted that if we didn t take this absolutely algorithmic approach, we could ve seen that 047 =, so. Thus every power of caluclated above could ve been reduced mod, to make the calculations easier. 9. (Niven.4.0 &.4. (a Suppose that n is a pseudoprime for the base a, but is not a strong pseudoprime. Show that there is then some k such that a k m ± (mod n but a k

6 (mod n. Prove that at least one of (n, m+ and (n, m is a nontrivial divisor of n. When the strong pseudoprime test fails and the pseudoprime test succeeds, we know we are exactly in the case where a n n i (mod n, but a i+ ± (mod n. This means that, letting k = n i+, ak m ± (mod n, but a k. Therefore m for some m ± (mod n. Then we have n (m = n (m + (m, but n (m + and n (m. If (n, m + =, then n (m + (m = n (m, which is a contradiction. If (n, m + = n, then n (m +, also a contradiction. Therefore (n, m + is a nontrivial divisor of n. (b Show that 34 is a pseudoprime for the base, but is not a strong pseudoprime. In particular, 85 m ± (mod 34, but 70 (mod 34. Find a nontrivial divisor of 34. So we have that 340 = 85, so we find 85 (mod 34. n n (mod Now, 85 = (mod 34. But 70 3 = 04 (mod 34. So we have m = 3. This gives us the possible prime divisors 3, 3,, and a quick division shows 34 = (Niven.4.4abd Use the Pollard rho method to find a proper divisor of (a 83, Starting with 3, and using the polynomial f(x = x +, we get the series: 3, 0, 0, 07, 4005, 5694,... Now, checking the gcd s of the differences of terms s and s, we have: (83,0-3=, (83,07-0=, but (83,5694-0=47. So we get 83 = (b 793, Here let s start with using the polynomial f(x = x, just to change things up. We get:, 3, 8, 63, 3968, 6066, 905, 3985, 6746, 85,... Now checking gcd s: (793,3-=, (793,63-3=, (793,6066-8=, (793, =, but (793, =4. Thus 793 = 4 93.

7 (c 609. This one is done exactly as the previous; using f(x = x + and beginning with, we find that (x 0 x 0, 609 = 83.. (Niven.5. Suppose that b a 67 (mod 9, with (a, 9 =. Find k such that b k a (mod 9. If b = 53, what is a mod 9? First, calculate φ(9 = 7. Next, use the Euclidean algorithm to find that =. Reducing mod 7 implies that 9 67equiv43 67 (mod 7, so k = 43. Using successive squaring then gives a (mod 9. (Bonus Define the composite factorials as (this is nonstandard notation n := m n m. (m,n= So p = (p!, and for example, 5 = (mod 5. (a Prove that if n = pq with p, q prime, then n (mod n. (b Determine a general formula for n mod n.