MATH 313: SOLUTIONS ASSIGNMENT 1. Problem 1
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1 MATH 1: SOLUTIONS ASSIGNMENT 1. Problem 1 Let be odd rime numbers such that + 4a for some nonzero integer a. Then (mod 4, a, otherwise. By the law of uadratic recirocity (Theorem 11.7, we have: ( ( Case 1: 1 (mod 4: ( ( 1 if either 1 (mod 4 or 1 (mod 4, 1 if (mod 4 1 ( Since + 4a, we have 4a, hence: ( a ( a Case 2: (mod 4: ( ( ( a ( ( a 1 a Since + 4a, we have 4a, hence: Problem 2 ( a ( a ( ( 4a ( ( ( + 4a ( ( ( 4a 4a ( ( ( ( ( ( a ( ( a 1 ( 4a a ( + 4a ( ( ( ( ( 4a 4a. First, note that gcd(, 10 1, so 2, 5. By the law of uadratic recirocity, since 5 1 (mod 4, we have ( ( 5. 5 Moreover, from class we know that:
2 ( 5 1 if ±1 (mod 5 1 if 2, (mod 5. Further, by Theorem 11.5, ( 1 Combining these two results, we conclude that : 1 if 1 (mod 4 1 if (mod 4. 1 if 1 (mod 4 ±1 (mod 5 ( ( ( if (mod 4 2, (mod 5 1 if 1 (mod 4 2, (mod 5 1 if (mod 4 ±1 (mod 5. We have 8 systems of congruences to solve, from which we obtain the following result: Problem ( 5 1 if 1,, 7, 9 (mod 20 1 if 11, 1, 17, 19 (mod 20. We will use the Chinese Remainder Theorem (CRT : if gcd(, 1, then the system x b (mod x c (mod has exactly 1 solution modulo. Now consider x 2 b (mod. We know that it has at most 2 solutions modulo, x, x. Case 1: If a, then x 2 a 0 (mod hence there is only one solution. Case 2: If 2 a, then x 2 a 1 (mod 2 so that there is one solution. Case : If odd a, then x 2 a (mod has either 0 or 2 solutions. Keeing those cases in mind, we can find the number of solutions for each of the two arts of the system: x 2 b (mod x 2 c (mod. Then deending on the number of solutions for each we find 0, 1, 2, or 4 linear systems to solve, each leading to a uniue solution. For instance, if we get 2 for each, such that we have 4 distinct solutions x, x, x, x, creating four different system of congruences of the form : x x (mod x x (mod
3 Note that if, we get different results since we are looking at modulo 2, we cannot use the CRT. Case 1: If 2 a, then x 2 a 0 (mod 2. There is only one solution. Case 2: If a but 2 a, then there is no solution. Case : If 2 a, then x ±1 (mod 2 only if 1 (mod 4. Case 4: If odd a, then x 2 a (mod 2 has either 0 or 2 solutions. Here the ossible number of solutions when is 0, 1, or 2 solutions. Problem 4 To solve x 2 x 1 0 (mod 1957, we begin by multilying by 4, so that we can comlete the suare to find that (2x (mod We thus want to find when 1 is a suare modulo Using the Legendre symbol the fact that (mod 4, we thus have ( ( ( ( ( hence there is a solution to our initial congruence. Problem 5 Claim 1: There are infinitely many rimes congruent to 2 modulo. Proof (by mimicking Euclid s roof : Suose that there are finitely many rimes congruent to 2 modulo such that they can be listed 0, 1,..., n, with 0 2. Now let N 1 n + 2. Observe that N is odd since the roduct of odd numbers is odd; thus 2 N. Note also that N 2 (mod ; thus N. Moreover, no odd rime congruent to 2 modulo (none of the 1,..., n divides N, otherwise they would have to divide 2 which is not ossible. But if doesn t divide N, no rime congruent to 2 mod divides N, then all rime divisors of N must be congruent to 1 mod. We have reach a contradiction, since if all rimes divisors of N are 1 mod, then N can be written as the roduct of those rimes thus would be 1 mod as well. But we have shown that N is congruent to 2 mod. The resulting contradiction imose that there are necessarily infinitely many rimes congruent to 2 mod. Claim 2: There are infinitely many rimes congruent to 1 modulo. Proof (by considering N ( k 2 + using Legendre symbols : Suose that there are finitely many rimes congruent to 1 mod such that they can be listed 1, 2,..., k. Now, let N ( k 2 +, let be any rime divisor of N. Note that cannot be 2 since N is mod 4, cannot be either, since N is 1 mod. We have ( k 2 (mod,
4 hence ( 1 whereby either ( ( 1 1 or In the first case, we have 1 (mod 4 so ( 1 ( 1 ( ( 1. imlies that 1 (mod. In the second case, necessarily (mod 4 so ( ( 1, we have that, again, 1 (mod. Therefore, in either case, we need 1 (mod ; that is any rime divisors of N must be 1 mod. This is a contradiction since by construction, any rime congruent to 1 mod cannot divide N. It follows that our initial assumtion must be false hence that there exist infinitely many rimes congruent to 1 mod. Problem 6 Observe that Let s first show that is a Carmichael number. Let a be an arbitrary integer such that gcd(a, , so that gcd(a, 7 gcd(a, 1 gcd(a, 7 1. Now consider a modulo 7, 1 7. We can check that 6, 0, 72 all divide Thus, by Fermat s little theorem, we know that a is 1 mod 7, mod 1, mod 7. Finally, by Chinese remainder Theorem, we find that a (mod Therefore, is a Carmichael number. Now let s show that is an euler seudorime to base 2. We want to comute the Jacobi symbol ( , which is: ( ( ( ( since all Legendre symbols are eual to 1. Now we need to show that (mod Since 7920 is divisible by 6,0, 72, by Fermat s little theorem, we get that a 7920 is 1 mod 7, mod 1, mod 7. Finally, by the Chinese Remainder Theorem, a (mod Finally, we need to show that is a strong seudorime to the base 2. So we write We need to check that either a (mod or a 495 2r 1 (mod Let s start by comuting mod Note that by definition of a seudorime, if (mod or (mod we are done. We can do the comutation as follows:
5 2 495 ( (mod ( (mod ( (mod 7 By Chinese Remier Theorem, we find that strong seudorime to the base 2. 1 (mod 15841, showing that is a Problem 7 Suose n is an Euler seudorime to the base, whereby we have (n 1/2 ( n (mod n. Moreover, if n 5 (mod 12, we have that n 1 (mod 4 n 2 (mod, hence ( n ( n ( 2 1, so (n 1/2 1 (mod n. It thus follows form the definition that n is a strong seudorime modulo.
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