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MATH 313: SOLUTIONS ASSIGNMENT 1. Problem 1 Let p and q be odd prime numbers such that p = q + 4a for some nonzero integer a. Then p ≡ q (mod 4), and p - a, otherwise p = q. By the law of quadratic reciprocity (Theorem 11.7), we have: ( p q 1 = q p −1 if either p ≡ 1 (mod 4) or q ≡ 1 (mod 4), if p ≡ q ≡ 3 (mod 4) Case 1: p ≡ q ≡ 1 (mod 4): p q p q −1 −1 =1⇒ = and = = 1. q p q p p q Since p = q + 4a, we have q = p − 4a, hence: a 4 a 4a q + 4a p = = = = q q q q q q and a 4 a −1 4a −4a p − 4a q = = = = = p p p p p p p p Case 2: p ≡ q ≡ 3 (mod 4): p q −1 −1 p q = −1 ⇒ =− and = = −1. q p q p p q Since p = q + 4a, we have q = p − 4a, hence: a 4 a 4a q + 4a p q = = = = =− q q q q q q p and a 4 a −1 4a −4a p − 4a q = =− =− =− =− . p p p p p p p p Problem 2 First, note that gcd(p, 10) = 1, so p 6= 2, 5. By the law of quadratic reciprocity, since 5 ≡ 1 (mod 4), we have 5 p = . p 5 Moreover, from class we know that: ( 5 1 = p −1 if p ≡ ±1 (mod 5) if p ≡ 2, 3 (mod 5). Further, by Theorem 11.5, ( −1 1 = p −1 if p ≡ 1 (mod 4) if p ≡ 3 (mod 4). Combining these two results, we conclude that : 1 1 −5 −1 5 = = p p p −1 −1 if p ≡ 1 (mod 4) and p ≡ ±1 (mod 5) if p ≡ 3 (mod 4) and p ≡ 2, 3 (mod 5) if p ≡ 1 (mod 4) and p ≡ 2, 3 (mod 5) if p ≡ 3 (mod 4) and p ≡ ±1 (mod 5). We have 8 systems of congruences to solve, from which we obtain the following result: ( −5 1 = p −1 if p ≡ 1, 3, 7, 9 (mod 20) if p ≡ 11, 13, 17, 19 (mod 20). Problem 3 We will use the Chinese Remainder Theorem (CRT) : if gcd(p, q) = 1, then the system ( x ≡ b (mod p) x ≡ c (mod q) has exactly 1 solution modulo pq. Now consider x2 ≡ b0 (mod p). We know that it has at most 2 solutions modulo p, xp , and x0p . Case 1: If p | a, then x2 ≡ a ≡ 0 (mod p) and hence there is only one solution. Case 2: If p = 2 and p - a, then x2 ≡ a ≡ 1 (mod 2) so that there is one solution. Case 3: If p odd and p - a, then x2 ≡ a (mod p) has either 0 or 2 solutions. Keeping those cases in mind, we can find the number of solutions for each of the two parts of the system: x2 ≡ b0 (mod p) and x2 ≡ c0 (mod q). Then depending on the number of solutions for each we find 0, 1, 2, or 4 linear systems to solve, each leading to a unique solution. For instance, if we get 2 for each, such that we have 4 distinct solutions x0p , x00p , x0q , x00q , creating four different system of congruences of the form : ( x ≡ xp (mod p) x ≡ xq (mod q) Note that if p = q, we get different results since we are looking at modulo p2 , and we cannot use the CRT. Case 1: Case 2: Case 3: Case 4: If p2 | a, then x2 ≡ a ≡ 0 (mod p2 ). There is only one solution. If p | a but p2 - a, then there is no solution. If p = 2 and p - a, then x ≡ ±1 (mod p2 ) only if p ≡ 1 (mod 4). If p odd and p - a, then x2 ≡ a (mod p2 ) has either 0 or 2 solutions. Here the possible number of solutions when p = q is 0, 1, or 2 solutions. Problem 4 To solve x2 − 3x − 1 ≡ 0 (mod 31957), we begin by multiplying by 4, so that we can complete the square to find that (2x − 3)2 − 13 ≡ 0 (mod 31957). We thus want to find when 13 is a square modulo 31957. Using the Legendre symbol and the fact that 31957 ≡ 1 (mod 4), we thus have 13 31957 3 13 1 = = = = =1 31957 13 13 3 3 and hence there is a solution to our initial congruence. Problem 5 Claim 1: There are infinitely many primes congruent to 2 modulo 3. Proof (by mimicking Euclid’s proof) : Suppose that there are finitely many primes congruent to 2 modulo 3 such that they can be listed p0 , p1 , ..., pn , with p0 = 2. Now let N = 3 · p1 · · · pn + 2. Observe that N is odd since the product of odd numbers is odd; thus 2 - N . Note also that N ≡ 2 (mod 3); thus 3 - N . Moreover, no odd prime congruent to 2 modulo 3 (none of the p1 , ..., pn ) divides N , otherwise they would have to divide 2 which is not possible. But if 3 doesn’t divide N , and no prime congruent to 2 mod 3 divides N , then all prime divisors of N must be congruent to 1 mod 3. We have reach a contradiction, since if all primes divisors of N are 1 mod 3, then N can be written as the product of those primes and thus would be 1 mod 3 as well. But we have shown that N is congruent to 2 mod 3. The resulting contradiction impose that there are necessarily infinitely many primes congruent to 2 mod 3. Claim 2: There are infinitely many primes congruent to 1 modulo 3. Proof (by considering N = (2p1 p2 ...pk )2 + 3 and using Legendre symbols) : Suppose that there are finitely many primes congruent to 1 mod 3 such that they can be listed p1 , p2 , ..., pk . Now, let N = (2p1 p2 ...pk )2 + 3, and let q be any prime divisor of N . Note that q cannot be 2 since N is 3 mod 4, and cannot be 3 either, since N is 1 mod 3. We have (2p1 p2 ...pk )2 ≡ −3 (mod q), and hence −3 q = 1 whereby either −1 3 −1 3 = = 1 or = = −1. q q q q In the first case, we have q ≡ 1 (mod 4) and so 3 q 1= = q 3 implies that q ≡ 1 (mod 3). In the second case, necessarily q ≡ 3 (mod 4) and so 3 q −1 = =− , q 3 and we have that, again, q ≡ 1 (mod 3). Therefore, in either case, we need q ≡ 1 (mod 3); that is any prime divisors of N must be 1 mod 3. This is a contradiction since by construction, any prime congruent to 1 mod 3 cannot divide N . It follows that our initial assumption must be false and hence that there exist infinitely many primes congruent to 1 mod 3. Problem 6 Observe that 15841 = 7 · 31 · 73. Let’s first show that 15841 is a Carmichael number. Let a be an arbitrary integer such that gcd(a, 15841) = 1, so that gcd(a, 7) = gcd(a, 31) = gcd(a, 73) = 1. Now consider a15840 modulo 7, 31 and 73. We can check that 6, 30, and 72 all divide 15840. Thus, by Fermat’s little theorem, we know that a15840 is 1 mod 7, mod 31, and mod 73. Finally, by Chinese remainder Theorem, we find that a15840 ≡ 1 (mod 15841). Therefore, 15840 is a Carmichael number. Now let’s show that 15841 is an euler pseudoprime to base 2. We want to compute the Jacobi 2 symbol 15841 , which is: 2 2 2 2 = =1 15841 7 31 73 since all 3 Legendre symbols are equal to 1. Now we need to show that 2 15841−1 2 = 27920 ≡ 1 (mod 15841). Since 7920 is divisible by 6,30, and 72, by Fermat’s little theorem, we get that a7920 is 1 mod 7, mod 31, and mod 73. Finally, by the Chinese Remainder Theorem, a7920 ≡ 1 (mod 15841). Finally, we need to show that 15841 is a strong pseudoprime to the base 2. So we write 15840 = 25 · r 495. We need to check that either a495 ≡ 1 (mod 15841) or a495·2 ≡ −1 (mod 15841). Let’s start by computing 2495 mod 15841. Note that by definition of a pseudoprime, if 2495 ≡ 1 (mod 15841) or 2495 ≡ −1 (mod 15841) we are done. We can do the computation as follows: 2495 ≡ (23 )165 ≡ 1 (mod 7) 2495 ≡ (25 )99 ≡ 1 (mod 31) 2495 ≡ (29 )55 ≡ 1 (mod 73) By Chinese Remiander Theorem, we find that 2495 ≡ 1 (mod 15841), showing that 15841 is a strong pseudoprime to the base 2. Problem 7 Suppose n is an Euler pseudoprime to the base 3, whereby we have 3(n−1)/2 ≡ n3 (mod n). Moreover, if n ≡ 5 (mod 12), we have that n ≡ 1 (mod 4) and n ≡ 2 (mod 3), and hence 3 n 2 = = = −1, n 3 3 and so 3(n−1)/2 ≡ −1 (mod n). It thus follows form the definition that n is a strong pseudoprime modulo 3.