# Math 4400/6400 Homework #8 solutions. 1. Let P be an odd integer (not necessarily prime). Show that modulo 2,

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1 MATH 4400 roblems. Math 4400/6400 Homework # solutions 1. Let P be an odd integer not necessarily rime. Show that modulo, { P 1 0 if P 1, 7 mod, 1 if P 3, mod. Proof. Suose that P 1 mod. Then we can write P k + 1, and so P 1 64k +16k+1 1 4k + k is even. Similarly, if P 7 mod, then we can write P k 1, and so P 1 64k 16k+1 1 4k k. If P 3 mod, then we can write P k + 3, and in this case P 1 4k + 3k + 1 is odd. Finally, when P mod, we can write P k 3, and in this case P 1 4k 3k + 1 is also odd.. Let P, Q be any odd integers. Show that modulo, P 1 Q 1 { 0 if either P 1 mod 4 or Q 1 mod 4, 1 if P 3 mod 4 and Q 3 mod 4. Proof. If P 1 mod 4, then we can write P 4k + 1, so that P 1 k is even, making the roduct P 1 Q 1 also even. The same reasoning shows that the roduct is even if Q 1 mod 4. So suose that both P and Q are congruent to 3 modulo 4. Then we can write P 4k + 3 and Q 4l + 3, so that P 1 k + 1 and Q 1 l + 1 are odd. Hence, P 1 Q 1 is also odd. 3. Use Gauss s lemma to rove that for each odd rime, { 1 if 1, 3 mod, 1 if, 7 mod. Proof. First we rove this in the case when 1 mod 4. Write 4k + 1, so that 1 k. Then each of the roducts 1,,..., k belongs to the interval /, 0. Shifting each of them forward by which leaves them unchanged modulo, we see they are congruent to integers from the interval /,. So if we write j ɛ j r j, with each ɛ j {±1} and each r j 0, /, then ɛ j 1 for all j with 1 j k. Now consider the remaining roducts k+1, k +,..., k. All of these belong to the interval, /, and so are congruent modulo to integers from the interval 0, /. Hence, ɛ j 1 for k + 1 j k. In total, there are k signs which are 1, and so by Gauss s lemma, 1 k. Thus, 1 if k is even and 1 if k is odd. The even k case corresonds to 1 mod, while the odd k case corresonds to mod. Now consider the case when 3 mod 4. Write 4k + 3, so that 1 k + 1. Then each of the roducts 1,,..., k belongs to the interval /, 0,

2 while each of the roducts k+1,..., k+1 belongs to, /. Reasoning as before, we see that ɛ j 1 for 1 j k and that ɛ j 1 for k + 1 j k + 1. Again, the total number of 1s is k, and so 1 k. Thus, 1 if k is even and 1 if k is odd. The even k case corresonds to 3 mod, and the odd k case corresonds to 7 mod. 4. Let m be a ositive integer. By definition, there are φm integers in [0, m that are relatively rime to m. Suose we list these out, say 0 a 1 < a < a 3 < < a φm < m. Show that every rime either divides m or is congruent, modulo m, to one of the a i. Proof. Let be a rime number. If does not divide m, then is relatively rime to m. Let a be the remainder when is divided by m, so that 0 a < m. Then a mod m. Moreover, you will remember from our roof of the Euclidean algorithm that and a have the same greatest common divisor with m. Thus gcda, m 1, and so a must be one of the a i.. Determine whether or not there is an integer x satisfying the congruence x 116 mod 09. Hint: 09 is a rime number. Proof. We have to comute We start off by noting that Now we aly quadratic recirocity. Since both 9 and 09 are rime, and 9 is 1 modulo 4, we have Now 3 and 9 are rime, and 9 is 1 modulo 4, so Now we know 3 1, since 3 7 mod. We also know from an examle done in class that 3 3 1, since 3 11 mod 1. Hence, Tracing through the above reasoning shows that , so that the congruence x 116 mod 09 does have a solution. In fact, the smallest ositive solution is given by x 376; this would not have been easy to find by hand! 6. Use the law of quadratic recirocity to determine the set of odd rimes for which 1. Exress your answer in terms of ossible residue classes of modulo 40. Solution. We ignore in what follows, since 0. Notice that. Thus, 1 if and only if 1 OR 1. As we showed in class, { 1 if 1, 7 mod, 1 if 3, mod.

3 On the other hand, since by QR, and since the quadratic residues modulo are 1, 4, we see that { 1 if 1, 4 mod, 1 if, 3 mod. We comuted this examle in class. Thus, if we want 1, we are led to solve the following four systems of congruences: a 1 mod and 1 mod, b 1 mod and 4 mod, c 7 mod and 1 mod, d 7 mod and 4 mod. By the Chinese remainder theorem, each of these has solutions, and the solution to each is unique modulo 40. Solving the systems, we find that the solutions are resectively 1 mod 40, 9 mod 40, 31 mod 40, and 39 mod 40. On the other hand, if we want 1, we are led to solve the following four systems of congruences: a 3 mod and mod, b 3 mod and 3 mod, c mod and mod, d mod and 3 mod. Solving these systems, we get the solutions 7 mod 40, 3 mod 40, 37 mod 40, and 13 mod 40. Putting everything together, we see that the odd rimes for which 1 are recisely those which satisfy 1, 3, 9, 13, 7, 31, 37, 39 mod This exercise is a continuation of one on your last HW assignment. Let F n n + 1 be the nth Fermat number. For this exercise, we suose that n. a Let be a rime dividing F n. Use the exercise from your revious HW to show that 1 mod. Proof. According to the revious week s HW, we have that 1 mod n+1. Since we are assuming that n, we have n + 1 3, and so n+1 1. Thus, 1 mod. b Exlain why the order of modulo divides 1. Proof. Since 1 mod, we know that is a quadratic residue modulo. So by Euler s criterion, 1/ 1 mod. Thus, the order of modulo must divide 1. 3

4 c Show that 1 mod n+. This is stronger than what you concluded in your last homework, where the modulus of the congruence was n+1. Proof. In last week s HW, you determined that the order of modulo was exactly n+1. Since the order divides 1 by art b, we see that 1 n+1 q for some q Z. Hence, 1 n+ q, and so 1 mod n+. 4

6 Proof. Suose that 3 Fn 1 1 mod F n. Let K Fn 1. Then 3 K 1 mod F n, so that 3 K 1 mod F n. Hence, the order of 3 modulo F n divides K but not K. However, K n 1 is a ower of, and so the only ositive integer dividing K but not K is K. So 3 has order K F n 1 modulo F n. Thus, F n 1 divides the size of the unit grou mod F n. But the size of the unit grou mod F n is smaller than F n 1 unless F n is rime. This roves the direction. For the direction, we first determine F n modulo 1. Notice that F 1 mod 1. But whenever F n mod 1, we have F n+1 F n mod 1. So by induction, each F n mod 1. So if F n is rime, then 3 F n 1. By Euler s criterion, 3 Fn mod F n. F n 6

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