Math 261 Exam 2. November 7, The use of notes and books is NOT allowed.

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1 Math 261 Eam 2 ovember 7, 2018 The use of notes and books is OT allowed Eercise 1: Polynomials mod 691 (30 ts In this eercise, you may freely use the fact that 691 is rime Consider the olynomials f( and g( in (Z/691Z[ 1 (5 ts Check that g( f(, and find a olynomial h( such that f( g(h( 2 (5 ts State the law of quadratic recirocity 3 (16 ts Use Legendre symbols to rove that neither g( nor h( have any roots in Z/691Z 4 (4 ts What is the comlete factorization of f( in (Z/691Z[? Make sure to justify that your factors are irreducible Solution 1: 1 By erforming the Euclidean division of f( by g(, we find quotient and remainder 0 So we can (and must take h( Let and q be distinct, odd rimes Then q ( 1 1 q q 3 For g(, we have 7 We comute that by quadratic recirocity and since (mod since (mod 7 and since is clearly a square mod 7 1 since 7 1 (mod 4 This shows that g( has not roots in Z/691Z 1

2 For h(, we have 13, and by quadratic recirocity and because (mod 13 1 since 13 ±3 (mod 8 So h( has no root either 4 If a olynomial of degree 2 is reducible, then it has a factor of degree 1, so it has a root Therefore g( and h( are irreducible, and is the comlete factorization of f( f( g(h( Eercise 2: The Péin test (30 ts In the 17th century, the French mathematician Pierre de Fermat studied the numbers F n 2 2n + 1, where n The urose of this eercise is to establish a criterion to test whether F n is rime In the rest of the eercise, we fi n 1 (4 ts Prove that F n 1 (mod 3 and that F n 1 (mod 4 ( 3 2 (4 ts Let be a rime such that 1 (mod 4 Prove that 3 (7 ts Use the revious questions to rove that if F n is rime, then 3 (Fn 1/2 1 (mod F n ( 3 4 (15 ts Conversely, rove that if 3 (Fn 1/2 1 (mod F n, then F n is rime Hint: Square both sides What is F n 1, and what does this tell you about the multilicative order of 3 mod F n? Solution 2: 1 Since 2 1 (mod 3, we have F n 2 2n + 1 ( 1 2n (mod 3 as n 1 Besides, 2 n so 2 2n is a multile of 4, whence F n 2 2n 1 (mod 4 2 This is an immediate consequence of quadratic recirocity 3 If F n is rime, then we have 3 (Fn 1/ (mod, and ( 3 by the revious question since clearly Fn 1 (mod 4 Since F n 1 (mod 3, ( , whence the result 2

3 4 If 3 (Fn 1/2 1 (mod F n, then 3 Fn 1 ( (mod F n, so the multilicative order of 3 mod F n divides F n 1 2 2n, which is a ower of 2 Since 3 (Fn 1/2 1 1 (mod F n, and since 2 is the only rime dividing F n 1, this order is in fact eactly F n 1 So the owers of 3 give us F n 1 elements in (Z/F n Z But the number of elements in (Z/F n Z is at most F n 1 since 0 is not invertible, so the owers of 3 give us all of (Z/F n Z (ie 3 is a rimitive root mod F n and all nonzero elements in Z/F n Z are invertible This means that Z/F n Z is a field, which imlies that F n is rime Remark: Fermat noticed that F n is rime for n 4, and claimed that F n was actually rime for every n Euler later showed that this was not true, since for instance F The rimality test resented in this eercise is named after the 19th century French mathematician Théohile Péin It only alies to Fermat numbers, but is much faster than the general-urose tests that can deal with any integer It was used in 1999 to rove that F 24 is comosite, which is quite an imressive feat since F 24 has digits! It is know today that F n is comosite for all 5 n 32 Proving that Fermat was totally wrong, in other words that F n is never rime when n 5, is still an oen roblem; in fact, as of today it is not known whether the digit number F 33 is rime Eercise 3: The Solovay-Strassen test (40 ts In this eercise, we fi an odd integer 3, not necessarily rime Let i1 v i i be its factorization, where the i are distinct rimes We define the Jacobi symbol by the formula vi Z ( for all Z, where i if is rime, then ( i1 i is the usual Legendre symbol defined in class In articular, 1 In this question, we investigate some basic roerties of the symbol The sub-questions of this question are indeendent from each other (a (3ts Prove that if y (mod, then [ [ y (b (6 ts Prove that [ 0 is invertible mod (c (3 ts Prove that [ [ y [ y We now introduce the function S : (Z/Z (Z/Z [ (10 ts Prove that if is rime, then S( 1 for all (Z/Z 3

4 3 The goal of this question is to rove that conversely, if is not rime, then S( is not always 1 In order to make things easier, we will suose that is comosite and squarefree, that is to say that 1 2 r with the i distinct rimes and r 2 We define M / 1 2 r t (a (10 ts Prove that there eists a t Z such that 1 1 and that t 1 (mod M Hint: CRT (b (8 ts Prove that if t is as in the revious question, then S(t 1 Hint: Comute S(t mod M Solution 3: 1 (a ( If ( y (mod, then y (mod i for all i, and therefore y i i for all i, so that [ [ y (b [ ( 0 i 0 for all i i for all i gcd(, 1 y ( vi y v i y vi vi y (c i1 i i1 i i i1 i i vi vi [ y i1 i i1 2 If is rime, then for all, [ ( is simly mod Therefore, if (Z/Z, then i by Fermat s little theorem S( , which is congruent to (a Since is odd, so is 1, so we know that in (Z/ 1 Z, half of the elements are not squares In articular, there eists t 1 (Z/ 1 Z such that t Let also t 2 1 Z/MZ ow 1 and M 2 r are corime since the i are airwise distinct, so by CRT we may find a t Z which reduces to t 1 mod 1 and to t 2 1 mod M This t solves the question t t 1 (b By construction we have 1 1 whereas i i +1 for all i 2, so [ t t i1 i Besides, since t 1 mod M, we obviously have t (mod M As a result, we have S(t (mod M If we had S(t 1 (mod, then we would also have S(t 1 (mod M, so that 1 1 (mod M Since M 2 r > 2, this is absurd, so S(t 1 (mod 4

5 Remark: The Jacobi symbol also obeys a form of quadratic recirocity, which makes it easy to evaluate even if the factorization of is unknown Besides, we clearly have is a square mod +1 for all (Z/Z ; however, the converse is not true, so the Jacobi symbol is less owerful that the Legendre symbol Once can rove that there eists t such that S(t 1 even when is not squarefree Besides, if such a t eists, then S( 1 for many Indeed, consider K { (Z/Z S( 1} One sees easily that for each k K, S(tk S(tS(k S(t 1 1; as a result, we can construct (at least #K elements tk (Z/Z such that S( 1, which shows that the roortion of (Z/Z such that S( 1 is 50% (In more technical terms, this is just saying that since S is a non-trivial grou homomorhism, its kernel has inde at least 2 Therefore, by trying a few at random, we will quickly find one such that S( 1, and so we will be able to detect that is not rime This test is thus useful to weed out comosite numbers, but it cannot be used to rove rigorously that a number is rime It has been suerseded by the Rabin-Miller test, which suffers from the same limitation but is more efficient ED 5

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