Solutions to Practice Problems


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1 Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2 4n ]. 2. There are φ(2 = 4 Dirichlet characters mod 2. A partial list of values is given by n χ 0 (n χ (n χ 2 (n χ 3 (n and χ i (n = for n = 2, 3, 4, 6, 8, 9, 0, 2, 4, 5, 6, 8, 20,... (where gcd(n, 2 >. 3. By the extended Euclidean Algorithm, find integers a, b such that am + k(p =. Let g r a mod p. By Fermat s Little Theorem, g m (r a m r am k r am (r p k r am+k(p = r = r mod p. Here we are assuming that gcd(r, p = ; if, however, r is divisible by p, then one may instead simply choose g = No; since x 2 y 2 0 or mod 4 for all integers x, y, we have x 2 3y 2 0, or 2 mod 4.
2 5. The primitive Pythagorean triples containing 00 are (2499, 00, 250 and (62, 00, 629. These can be found by any of our approaches to the classification of primitive Pythagorean triples. 6. Euclid s Lemma states that if a prime p divides a product ab of two integers a and b, then p divides at least one of the factors a, b. To prove this, we may suppose that p does not divide a, so that gcd(a, p =. By Euclid s Algorithm, it follows that ra + sp = for some integers r, s; but then p divides r(ab + (sbp = (ra + spb = b since it divides both terms on the left side. 7. A prime p has 00 digits iff it satisfies 0 99 < p < Such primes lie in the interval [, 0 00 ] but not in the interval [, 0 99 ]. The number of such primes is by the Prime Number Theorem. π(0 00 π( ln( ln( The nonzero elements of F p, except for and, occur in pairs {a, a }. So in the product 0 a F p a, all factors cancel (in pairs except for the factors and, leaving a value of as the final value of the product. (Of course when p = 2 there is really only one such factor since = in this case; but our answer holds since the final product is once again =. 9. The Dirichlet character mod 2 defined by { 0, if n is even; χ(n =, if n is odd has Dirichlet series with Euler factorization given by L(χ, s = + 3 2s + 5 2s + 7 2s + 9 2s + 2s + = ( p χ(p p ( ( s ( ( = 3 s 5 s 7. s s 2
3 This is almost the same as the Euler factorization for the Riemann zetafunction; only the Euler factor for the prime p = 2 is missing, so ( 2 s L(χ, s = ζ(s. Evaluating at s = 2 gives 4 π2 3L(χ, 2 = ζ(2 = 6. Solving for the given series, we obtain L(χ, 2 = π Recall that n is perfect iff σ(n = 2n. Here σ is a multiplicative function satisfying σ(p r = + p + p p r + p r = p r p p r < pr p. If n = p r for some odd prime p and r, then we require 2n = σ(n < n p so 2 < p 3 = 3 2, a contradiction. If n = p r q s for distinct odd primes p, q with exponents r, s, then we require 2n = 2p r q s = σ(p r σ(q s < pr p q s q so 2 < again, this is impossible. ( ( p q ( 3 ( 5 = 5 8 ;. No; the only Mersenne prime with last decimal digit equal to 3 is M 2 = 2 2 = 3. Let M p = 2 p where p is prime. If p = 4k + for some integer k, then M p = 2 4k+ = 2(6 k 2 6 mod 0; whereas if p = 4k + 3 then M p = 2 4k+3 = 8(6 k mod 0. 3
4 2. If a is relatively prime with 05 = 5 3 7, then by Fermat s Little Theorem a 04 = (a mod 5; a 04 = (a mod 3; and a 04 = (a mod 7 so a mod 05. (This last conclusion follows from the Chinese Remainder Theorem; but it is a more direct consequence of the fact that a is divisible by three primes 5, 3 and 7, so it is divisible by their product. 3. By Fermat s Little Theorem, (M /2 (a 2 (M /2 a M mod M, so (M+/2 (M /2. One solution is given by a (M+/4 mod M. (Since M + is divisible by 4, the exponent (M + /4 is an integer. The complete list of solutions is given by a ± (M+/4 mod M. 4. Let p, p 2,..., p r be the distinct primes dividing n; then we must solve n = 2φ(n = 2n ( p ( p 2 ( p r, i.e. = 2 ( p ( p 2 ( p r. If n has any odd prime divisors p i, then the right hand side is never an integer (its denominator will be divisible by p i. So we must have n = 2 k where k. 5. (a The primes having at most 50 decimal digits lie in the interval [, 0 50 ]. Their number is by the Prime Number Theorem. π( ln( (b The primes having exactly 50 decimal digits lie in the interval [0 49, 0 50 ]. Their number is π(0 50 π( ln( ln( (c Large primes end in either, 3, 7 or 9, in roughly equal numbers; so the number of 50digit primes ending in 3 is roughly 4( π(0 50 π(
5 (d Large primes end in either X, X3, X7 or X9, in roughly equal numbers, where X represents any decimal digit; so the number of 50digit primes ending in 33 is roughly 40( π(0 50 π( By the extended Euclidean algorithm (or, in this simple case, by quick inspection we find 2 5 =, so the inverse of 5 mod is 2 9 and 5x 9 mod x mod. Write x = k + 4. Now we must solve 6x 5k mod 6, i.e. 5k 42 mod 6. Once again, = so the inverse of 5 mod 6 is 2 49 and k mod 6. Writing k = 6t + 45, this gives x = k + 4 = 67t In other words, the simultaneous solutions of the original pair of linear congruences are given by x 499 mod No, it is not possible for f(n to be prime for every positive integer n. Since f(x is nonconstant, it is either increasing or decreasing for all sufficiently large x. In the decreasing case, f(x is eventually negative for large x, and therefore not prime; so we may assume there is a positive integer n such that f(x is increasing for x n. By assumption, p = f(n is prime. Also f(n+p f(n p 0 mod p, i.e. f(n+p is divisible by p. Since f(n+p is also prime, we must have f(n+p = p. This contradicts f(n+p > f(n. 8. Yes. If p is composite, then there exists a {2, 3,..., p } dividing p; but then gcd(a p, p a >. (Note: Carmichael numbers are no exception to this fact. In order for n to be a Carmichael number, one only requires a n mod n for every a relatively prime to n. 5
6 Solutions to Practice True/False Questions F F F F T F T T F F Although you are not required to explain your answers to True/False questions, the following remarks may help to understand the solution key.. This special case of Fermat s Last Theorem was proved many decades before the full theorem was finally proved. 2. Fermat s Little Theorem is used to verify that certain numbers are composite; but it is of no direct value in finding prime factors of integers. 3. N 2 = (N + (N is prime only for N { 2, 2}. 4. A counterexample is given by = By Dirichlet s Theorem, there are infinitely many primes congruent to 7 mod 8. None of these are expressible as a sum of three squares. 6. Consider that 6 divides 4 9 = 36, but 6 divides neither 4 nor 9. (This shows that in the statement of Euclid s Lemma, the hypothesis that p is prime is essential. 7. This fact, which is easily proved using the Fundamental Theorem of Arithmetic, was used in our classification of primitive Pythagorean triples. 8. Recall that σ(p is the sum of all positive divisors of p, including and p itself; thus σ(p = + p + ( sum of divisors of p between and p. The latter sum is zero iff p is prime. 9. The only prime p for which p 2 + is prime, is p = 2. All larger values of the prime p yield an even value of p A counterexample is given by r = s = a = b =. From the relation ra + sb = d, one can conlcude is that gcd(a, b divides d, not that gcd(a, b = d. 6
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