# Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

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1 Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations. We devote the next three chapters to answering the following types of questions: Is 3 congruent to the square of some number modulo 7? Does the congruence x 2 1 (mod 13) have a solution? For which primes p does the congruence x 2 2 (mod p) have a solution? We can answer the first two questions right now. To see if 3 is congruent to the square of some number modulo 7, we just square each of the numbers from 0 to 6, reduce modulo 7, and see if any of them are equal to 3. Thus, (mod 7) (mod 7) (mod 7) (mod 7) (mod 7) (mod 7) (mod 7). 5

2 [Chap. 23] Squares Modulo p 6 So we see that 3 is not congruent to a square modulo 7. In a similar fashion, if we square each number from 0 to 12 and reduce modulo 13, we find that the congruence x 2 1 (mod 13) has two solutions, x 5 (mod 13) and x 8 (mod 13). 1 As always, we need to look at some data before we can even begin to look for patterns and make conjectures. Here are some tables giving all the squares modulo p for p = 5, 7, 11, and 13. b b Modulo 5 b b Modulo 7 b b Modulo 11 b b Modulo 13 Many interesting patterns are already apparent from these lists. For example, each number (other than 0) that appears as a square seems to appear exactly twice. Thus, 5 is both 4 2 and 7 2 modulo 11, and 3 is both 4 2 and 9 2 modulo 13. In fact, if we fold each list over in the middle, the same numbers appear as squares on the top and on the bottom. How can we describe this pattern with a formula? We are saying that the square of the number b and the square of the number p b are the same modulo p. But 1 For many years during the nineteenth century, mathematicians were uneasy with the idea of the number 1. Its current appellation imaginary number still reflects that disquiet. But if you work modulo 13, for example, then there s nothing mysterious about 1. In fact, 5 and 8 are both square roots of 1 modulo 13. 6

4 [Chap. 23] Squares Modulo p 8 since the same numbers are repeated in reverse order if we square the remaining numbers ( ) p + 1 2,..., (p 2) 2, (p 1) 2 (mod p). 2 So in order to show that there are exactly (p 1)/2 quadratic residues, we need to check that the numbers 1 2, 2 2,..., ( p 1) 2 2 are all different modulo p. Suppose that b 1 and b 2 are numbers between 1 and (p 1)/2, and suppose that b 2 1 b2 2 (mod p). We want to show that b 1 = b 2. The fact that b 2 1 b2 2 (mod p) means that p divides b 2 1 b2 2 = (b 1 b 2 )(b 1 + b 2 ). However, b 1 + b 2 is between 2 and p 1, so it can t be divisible by p. Thus p must divide b 1 b 2. But b 1 b 2 < (p 1)/2, so the only way for b 1 b 2 to be divisible by p is to have b 1 = b 2. This shows that the numbers 1 2, 2 2,..., ( p 1) 2 2 are all different modulo p, so there are exactly (p 1)/2 quadratic residues modulo p. Now we need only observe that there are p 1 numbers between 1 and p 1, so if half of them are quadratic residues, the other half must be nonresidues. Suppose that we take two quadratic residues and multiply them together. Do we get a QR or an NR, or do we sometimes get one and sometimes the other? For example, 3 and 10 are QRs modulo 13, and their product 3 10 = 30 4 is again a QR modulo 13. Actually, this should have been clear without any computation, since if we multiply two squares, we should get a square. We can formally verify this in the following way. Suppose that a 1 and a 2 are both QRs modulo p. This means that there are numbers b 1 and b 2 so that a 1 b 2 1 (mod p) and a 2 b 2 2 (mod p). Multiplying these two congruences together, we find that a 1 a 2 (b 1 b 2 ) 2 (mod p), which shows that a 1 a 2 is a QR. The situation is less clear if we multiply a QR by an NR, or if we multiply two NRs together. Here are some examples using the data in our tables: QR NR?? (mod p) (mod 7) NR (mod 11) NR (mod 13) NR (mod 13) NR NR NR?? (mod p) (mod 7) QR (mod 11) QR (mod 13) QR (mod 13) QR Thus, multiplying a quadratic residue and a nonresidue seems to yield a nonresidue, while the product of two nonresidues always seems to be a residue. Symbolically, we might write QR QR = QR, QR NR = NR, NR NR = QR. 8

5 [Chap. 23] Squares Modulo p 9 We ve already seen that the first relation is true, and we now verify the other two relations. Theorem 23.2 (Quadratic Residue Multiplication Rule). (Version 1) Let p be an odd prime. Then: (i) The product of two quadratic residues modulo p is a quadratic residue. (ii) The product of a quadratic residue and a nonresidue is a nonresidue. (iii) The product of two nonresidues is a quadratic residue. These three rules can by summarized symbolically by the formulas QR QR = QR, QR NR = NR, NR NR = QR. Verification. We have already seen that QR QR = QR. Suppose next that a 1 is a QR, say a 1 b 2 1 (mod p), and that a 2 is an NR. We are going to assume that a 1 a 2 is a QR and derive a contradiction. The assumption that a 1 a 2 is a QR means that it is congruent to b 2 3 for some b 3, so we have b 2 3 a 1 a 2 b 2 1a 2 (mod p). Note that gcd(b 1, p) = 1, since p a 1 and a 1 = b 2 1, so the Linear Congruence Theorem (Theorem 8.1) says that we can find an inverse for b 1 modulo p. In other words, we can find some c 1 such that c 1 b 1 1 (mod p). Multiplying both sides of the above congruence by c 2 1 gives c 2 1b 2 3 c 2 1a 1 a 2 (c 1 b 1 ) 2 a 2 a 2 (mod p). Thus a 2 (c 1 b 3 ) 2 (mod p) is a QR, contradicting the fact that a 2 is a NR. This completes the proof that QR NR = NR. We are left to deal with the product of two NRs. Let a be an NR and consider the set of values a, 2a, 3a,..., (p 2)a, (p 1)a (mod p). By an argument we ve used before (see Claim 9.2 on page 63), these are just the numbers 1, 2,..., (p 1) rearranged in some different order. In particular, they include the 1 2 (p 1) QRs and the 1 2 (p 1) NRs. However, as we already proved, each time that we multiply a by a QR, we get an NR, so the 1 2 (p 1) products a QR 9

6 [Chap. 23] Squares Modulo p 10 already give us all 1 2 (p 1) QRs in the list. Hence when we multiply a by an NR, the only possibility is that it is equal to one of the NRs in the list, because the a QR products have already used up all of the QRs in the list. 2 This completes the verification of the quadratic residue multiplication rules. Now take a minute to stare at QR QR = QR, QR NR = NR, NR NR = QR. Do these rules remind you of anything? If not, here s a hint. Suppose that we try to replace the symbols QR and NR with numbers. What numbers would work? That s right, the symbol QR behaves like +1 and the symbol NR behaves like 1. Notice that the somewhat mysterious third rule, the one that says that the product of two nonresidues is a quadratic residue, reflects the equally mysterious rule 3 ( 1) ( 1) = +1. Having observed that QRs behave like +1 and NRs behave like 1, Adrien- Marie Legendre introduced the following useful notation. The Legendre symbol of a modulo p is ( ) { a 1 if a is a quadratic residue modulo p, = p 1 if a is a nonresidue modulo p. For example, data from our earlier tables says that ( ) ( ) ( ) = 1, = 1, = 1, ( ) 3 = 1. 7 Using the Legendre symbol, our quadratic residue multiplication rules can be given by a single formula. Theorem 23.3 (Quadratic Residue Multiplication Rule). (Version 2) Let p be an odd prime. Then ( )( ) ( ) a b ab =. p p p 2 When you have eliminated all of the quadratic residues, the remaining numbers, no matter how improbable, must be the nonresidues! (with apologies to Sherlock Holmes and Sir Arthur Conan Doyle). 3 You may no longer consider the formula ( 1) ( 1) = +1 mysterious, since it s so familiar to you. But you should have found it mysterious the first time you saw it. And if you stop to think about it, there is no obvious reason why the product of two negative numbers should equal a positive number. Can you come up with a convincing argument that ( 1) ( 1) must equal +1? 10

7 [Chap. 23] Squares Modulo p 11 The Legendre symbol is useful for making calculations. For example, suppose that we want to know if 75 is a square modulo 97. We can compute ( ) ( ) ( )( )( ) ( ) = = = Notice that it doesn t matter whether ( 5 97) is +1 or 1, since it appears twice, and (+1) 2 = ( 1) 2 = 1. Now we observe that (mod 97), so 3 is a QR. Hence, ( ) ( ) 75 3 = = Of course, we were lucky in being able to recognize 3 as a QR modulo 97. Is there some way to evaluate a Legendre symbol like ( 3 97) without relying on luck or trial and error? The answer is yes, but that s a topic for another chapter. Exercises Make a list of all the quadratic residues and all the nonresidues modulo Write a program that takes as input a prime p and produces as output the two numbers A = sum of all 1 a < p such that a is a quadratic residue modulo p, B = sum of all 1 a < p such that a is a nonresidue modulo p. For example, if p = 11, then the quadratic residues are (mod 11), (mod 11), (mod 11), (mod 11), (mod 11), so A = = 22 and B = = 33. (a) Make a list of A and B for all primes p < 100. (b) What is the value of A + B? Prove that your guess is correct. (c) Compute A mod p and B mod p. Find a pattern and prove that it is correct. (d) For which primes is it true that A = B? After reading Chapter 24, prove that your guess is correct. (e) If A B, which one tends to be larger, A or B? Try to prove that your guess is correct, but be forewarned that this is a very difficult problem A number a is called a cubic residue modulo p if it is congruent to a cube modulo p, that is, if there is a number b such that a b 3 (mod p). 11

8 [Chap. 23] Squares Modulo p 12 (a) Make a list of all the cubic residues modulo 5, modulo 7, modulo 11, and modulo 13. (b) Find two numbers a 1 and b 1 such that neither a 1 nor b 1 is a cubic residue modulo 19, but a 1 b 1 is a cubic residue modulo 19. Similarly, find two numbers a 2 and b 2 such that none of the three numbers a 2, b 2, or a 2 b 2 is a cubic residue modulo 19. (c) If p 2 (mod 3), make a conjecture as to which a s are cubic residues. Prove that your conjecture is correct. 12

10 [Chap. 24] Is 1 a Square Modulo p? Is 2? 14 Reading from this table, we compile the following data: 1 is a quadratic residue for p = 5, 13, 17, is a nonresidue for p = 3, 7, 11, 19, 23, 31. It s not hard to discern the pattern. If p is congruent to 1 modulo 4, then 1 seems to be a quadratic residue modulo p, and if p is congruent to 3 modulo 4, then 1 seems to be a nonresidue. We can express this guess using Legendre symbols, ( ) 1 p? = { 1 if p 1 (mod 4), 1 if p 3 (mod 4). Let s check our conjecture on the next few cases. The next two primes, 37 and 41, are both congruent to 1 modulo 4 and, sure enough, x 2 1 (mod 37) has the solutions x 6 and 31 (mod 37), and x 2 1 (mod 41) has the solutions x 9 and 32 (mod 41). Similarly, the next two primes 43 and 47 are congruent to 3 modulo 4, and we check that 1 is a nonresidue for 43 and 47. Our guess is looking good! The tool that we use to verify our conjecture might be called the Square Root of Fermat s Little Theorem. How, you may well ask, does one take the square root of a theorem? Recall that Fermat s Little Theorem (Chapter 9) says a p 1 1 (mod p). We won t really be taking the square root of this theorem, of course. Instead, we take the square root of the quantity a p 1 and ask for its value. So we want to answer the following question: Let A = a (p 1)/2. What is the value of A modulo p? One thing is obvious. If we square A, then Fermat s Little Theorem tells us that A 2 = a p 1 1 (mod p). Hence, p divides A 2 1 = (A 1)(A + 1), so either p divides A 1 or p divides A + 1. (Notice how we are using the property of prime numbers proved on page 44.) Thus A must be congruent to either +1 or 1. 14

11 [Chap. 24] Is 1 a Square Modulo p? Is 2? 15 Here are a few random values of p, a, and A. For comparison purposes, we have also included the value of the Legendre symbol ( a p). Do you see a pattern? p a A (mod p) ( a p) It certainly appears that A 1 (mod p) when a is a quadratic residue and that A 1 (mod p) when a is a nonresidue. In other words, it looks like A (mod p) has the same value as the Legendre symbol ( a p). We use a counting argument to verify this assertion, which goes by the name of Euler s Criterion. Theorem 24.1 (Euler s Criterion). Let p be an odd prime. Then ( ) a a (p 1)/2 (mod p). p Verification. Suppose first that a is a quadratic residue, say a b 2 (mod p). Then Fermat s Little Theorem (Theorem 9.1) tells us that a (p 1)/2 (b 2 ) (p 1)/2 = b p 1 1 (mod p). Hence a (p 1)/2 ( a p) (mod p), which is Euler s criterion when a is a quadratic residue. We next consider the congruence X (p 1)/2 1 0 (mod p). We have just proven that every quadratic residue is a solution to this congruence, and we know from Theorem 23.1 that there are exactly 1 2 (p 1) distinct quadratic residues. We also know from the Polynomial Roots Mod p Theorem (Theorem 8.2) that this polynomial congruence can have at most 1 2 (p 1) distinct solutions. Hence { solutions to X (p 1)/2 1 0 (mod p) } = { quadratic residues modulo p }. Now let a be a nonresidue. Fermat s Little Theorem tells us that a p 1 1 (mod p), so 0 a p 1 1 (a (p 1)/2 1)(a (p 1)/2 + 1) (mod p). 15

12 [Chap. 24] Is 1 a Square Modulo p? Is 2? 16 The first factor is not zero modulo p, because we already showed that the solutions to X (p 1)/2 1 0 (mod p) are the quadratic residues. Hence the second factor must vanish modulo p, so ( ) a a (p 1)/2 1 = (mod p). p This shows that Euler s criterion is also true for nonresidues. Using Euler s Criterion, it is very easy to determine if 1 is a quadratic residue modulo p. For example, if we want to know whether 1 is a square modulo the prime p = 6911, we just need to compute ( 1) (6911 1)/2 = ( 1) 3455 = 1. Euler s Criterion then tells us that ( ) 1 1 (mod 6911) But ( ) ( a p is always either +1 or 1, so in this case we must have ) = 1. Hence, 1 is a nonresidue modulo Similarly, for the prime p = 7817 we find that ( 1) (7817 1)/2 = ( 1) 3908 = 1. Hence, ( ) = 1, so 1 is a quadratic residue modulo Observe that, although we now know that the congruence x 2 1 (mod 7817) is solvable, we still don t have any efficient way to find a solution. The solutions turn out to be x 2564 (mod 7817) and x 5253 (mod 7817). As these two examples make clear, Euler s Criterion can be used to determine exactly which primes have 1 as a quadratic residue. This elegant result, which answers the initial question in the title of this chapter, is the first part of the Law of Quadratic Reciprocity. Theorem 24.2 (Quadratic Reciprocity). (Part I) Let p be an odd prime. Then 1 is a quadratic residue modulo p if p 1 (mod 4), and 1 is a nonresidue modulo p if p 3 (mod 4). In other words, using the Legendre symbol, ( ) { 1 1 if p 1 (mod 4), = p 1 if p 3 (mod 4). 16

13 [Chap. 24] Is 1 a Square Modulo p? Is 2? 17 Verification. Euler s Criterion says that ( 1) (p 1)/2 ( ) 1 (mod p). p Suppose first that p 1 (mod 4), say p = 4k + 1. Then ( ) 1 ( 1) (p 1)/2 = ( 1) 2k = 1, so 1 (mod p). p But ( ) 1 p is either +1 or 1, so it must equal 1. This proves that if p 1 (mod 4) then ( ) 1 p = 1. Next we suppose that p 3 (mod 4), say p = 4k + 3. Then ( ) 1 ( 1) (p 1)/2 = ( 1) 2k+1 = 1, so 1 (mod p). p This shows that ( ) 1 p must equal 1, which completes the verification of Quadratic Reciprocity (Part I). We can use the first part of quadratic reciprocity to answer a question left over from Chapter 12. As you may recall, we showed that there are infinitely many primes that are congruent to 3 modulo 4, but we left unanswered the analogous question for primes congruent to 1 modulo 4. Theorem 24.3 (Primes 1 (Mod 4) Theorem). There are infinitely many primes that are congruent to 1 modulo 4. Proof. Suppose we are given a list of primes p 1, p 2,..., p r, all of which are congruent to 1 modulo 4. We are going to find a new prime, not in our list, that is congruent to 1 modulo 4. Repeating this process gives a list of any desired length. Consider the number A = (2p 1 p 2 p r ) We know that A can be factored into a product of primes, say A = q 1 q 2 q s. It is clear that q 1, q 2,..., q s are not in our original list, since none of the p i s divide A. So all we need to do is show that at least one of the q i s is congruent to 1 modulo 4. In fact, we ll see that all of them are. 17

14 [Chap. 24] Is 1 a Square Modulo p? Is 2? 18 First we note that A is odd, so all the q i s are odd. Next, each q i divides A, so (2p 1 p 2 p r ) = A 0 (mod q i ). This means that x = 2p 1 p 2 p r is a solution to the congruence x 2 1 (mod q i ), so 1 is a quadratic residue modulo q i. Now Quadratic Reciprocity tells us that q i 1 (mod 4). We can use the procedure described in this proof to produce a list of primes that are congruent to 1 modulo 4. Thus, if we start with p 1 = 5, then we form A = (2p 1 ) = 101, so our second prime is p 2 = 101. Then A = (2p 1 p 2 ) = , which is again prime, so our third prime is p 3 = We ll go one more step, A = (2p 1 p 2 p 3 ) = = Notice that all the primes 53, 1613, and are congruent to 1 modulo 4, just as predicted by the theory. Having successfully answered the first question in the title of this chapter, we move on to the second question and consider a = 2, that oddest of all primes. Just as we did with a = 1, we are looking for some simple characterization for the primes p such that 2 is a quadratic residue modulo p. Can you find the pattern in the following data, where the line labeled x 2 2 gives the solutions to x 2 2 (mod p) if 2 is a quadratic residue modulo p and is marked NR if 2 is a nonresidue? p x 2 2 NR NR 3, 4 NR NR 6, 11 NR 5, 18 NR 8, 23 p x 2 2 NR 17, 24 NR 7, 40 NR NR NR NR 12, 59 32, 41 p x 2 2 9, 70 NR 25, 64 14, 83 NR 38, 65 NR NR 51, 62 16,

15 [Chap. 24] Is 1 a Square Modulo p? Is 2? 19 Here s the list of primes separated according to whether 2 is a residue or a nonresidue. 2 is a quadratic residue for p = 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, is a nonresidue for p = 3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109 For a = 1, it turned out that the congruence class of p modulo 4 was crucial. Is there a similar pattern if we reduce these two lists of primes modulo 4? Here s what happens if we do. 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3 (mod 4), 3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109 3, 1, 3, 1, 3, 1, 1, 3, 1, 3, 1, 3, 3, 1, 3, 1 (mod 4). This doesn t look too promising. Maybe we should try reducing modulo 3. 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1 (mod 3) 3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109 0, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1 (mod 3). This doesn t look any better. Let s make one more attempt before we give up. What happens if we reduce modulo 8? 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127 7, 1, 7, 7, 1, 7, 7, 1, 7, 1, 1, 7, 1, 7 (mod 8) 3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109 3, 5, 3, 5, 3, 5, 5, 3, 5, 3, 5, 3, 3, 5, 3, 5 (mod 8). Eureka! It surely can t be a coincidence that the first line is all 1 s and 7 s and the second line is all 3 s and 5 s. This suggests the general rule that 2 is a quadratic residue modulo p if p is congruent to 1 or 7 modulo 8 and that 2 is a nonresidue if p is congruent to 3 or 5 modulo 8. In terms of Legendre symbols, we would write ( ) 2 p? = { 1 if p 1 or 7 (mod 8), 1 if p 3 or 5 (mod 8). 19

16 [Chap. 24] Is 1 a Square Modulo p? Is 2? 20 Can we use Euler s Criterion to verify our guess? Unfortunately, the answer is no, or at least not in any obvious way, since there doesn t seem to be an easy method to calculate 2 (p 1)/2 (mod p). However, if you go back and examine our proof of Fermat s Little Theorem in Chapter 9, you ll see that we took the numbers 1, 2,..., p 1, multiplied each one by a, and then multiplied them all together. This gave us a factor of a p 1 to pull out. In order to use Euler s Criterion, we only want 1 2 (p 1) factors of a to pull out, so rather than starting with all the numbers from 1 to p, we just take the numbers from 1 to 1 2 (p 1). We illustrate this idea, which is due to Gauss, to determine if 2 is a quadratic residue modulo 13. We begin with half the numbers from 1 to 12: 1, 2, 3, 4, 5, 6. If we multiply each by 2 and then multiply them together, we get = (2 1)(2 2)(2 3)(2 4)(2 5)(2 6) = = 2 6 6!. Notice the factor of 2 6 = 2 (13 1)/2, which is the number we re really interested in. Gauss s idea is to take the numbers 2, 4, 6, 8, 10, 12 and reduce each of them modulo 13 to get a number lying between 6 and 6. The first three stay the same, but we need to subtract 13 from the last three to get them into this range. Thus, 2 2 (mod 13) 4 4 (mod 13) 6 6 (mod 13) 8 5 (mod 13) 10 3 (mod 13) 12 1 (mod 13). Multiplying these numbers together, we find that ( 5) ( 3) ( 1) ( 1) ! (mod 13). Equating these two values of (mod 13), we see that 2 6 6! 6! (mod 13). This implies that (mod 13), so Euler s Criterion tells us that 2 is a nonresidue modulo 13. Let s briefly use the same ideas to check if 2 is a quadratic residue modulo 17. We take the numbers from 1 to 8, multiply each by 2, multiply them together, and calculate the product in two different ways. The first way gives = 2 8 8!. 20

17 [Chap. 24] Is 1 a Square Modulo p? Is 2? 21 For the second way, we reduce modulo 17 to bring the numbers into the range from 8 to 8. Thus, 2 2 (mod 17) 4 4 (mod 17) 6 6 (mod 17) 8 8 (mod 17) 10 7 (mod 17) 12 5 (mod 17) 14 3 (mod 17) 16 1 (mod 17). Multiplying these together gives ( 7) ( 5) ( 3) ( 1) ( 1) 4 8! (mod 17). Therefore, 2 8 8! ( 1) 4 8! (mod 17), so (mod 17), and hence 2 is a quadratic residue modulo 17. Now let s think about Gauss s method a little more generally. Let p be any odd prime. To make our formulas simpler, we let P = p 1 2. We start with the even numbers 2, 4, 6,..., p 1. Multiplying them together and factoring out a 2 from each number gives (p 1) = 2 (p 1)/ p 1 2 = 2 P P!. The next step is to take the list 2, 4, 6,..., p 1 and reduce each number modulo p so that it lies in the range from P to P, that is, between (p 1)/2 and (p 1)/2. The first few numbers won t change, but at some point in the list we ll start hitting numbers that are larger than (p 1)/2, and each of these large numbers needs to have p subtracted from it. Notice that the number of minus signs introduced is exactly the number of times we need to subtract p. In other words, Number of integers in the list Number of minus signs = 2, 4, 6,..., (p 1). that are larger than 1 2 (p 1) The following illustration may help to explain this procedure } {{ } Numbers (p 1)/2 are left unchanged. Comparing the two products, we get (p 5) (p 3) (p 1) } {{ } Numbers > (p 1)/2. Need to substract p from each. 2 P P! = (p 1) ( 1) (Number of minus signs) P! (mod p), 21

18 [Chap. 24] Is 1 a Square Modulo p? Is 2? 22 so canceling P! from each side gives the fundamental formula 2 (p 1)/2 ( 1) (Number of minus signs) (mod p). Using this formula, it is easy to verify our earlier guess, thereby answering the second question in the chapter title. Theorem 24.4 (Quadratic Reciprocity). (Part II) Let p be an odd prime. Then 2 is a quadratic residue modulo p if p is congruent to 1 or 7 modulo 8, and 2 is a nonresidue modulo p if p is congruent to 3 or 5 modulo 8. In terms of the Legendre symbol, ( ) { 2 1 if p 1 or 7 (mod 8), = p 1 if p 3 or 5 (mod 8). Verification. There are actually four cases to consider, depending on the value of p (mod 8). We do two of them and leave the other two for you. We start with the case that p 3 (mod 8), say p = 8k + 3. We need to list the numbers 2, 4,..., p 1 and determine how many of them are larger than 1 2 (p 1). In this case, p 1 = 8k + 2 and 1 2 (p 1) = 4k + 1, so the cutoff is as indicated in the following diagram: k (4k + 2) (4k + 4) (8k + 2). We need to count how many numbers there are to the right of the vertical bar. In other words, how many even numbers are there between 4k + 2 and 8k + 2? The answer is 2k + 1. (If this isn t clear to you, try a few values for k and you ll see why it s correct.) This shows that there are 2k + 1 minus signs, so the fundamental formula given above tells us that 2 (p 1)/2 ( 1) 2k+1 1 (mod p). Now Euler s Criterion says that 2 is a nonresidue, so we have proved that 2 is a nonresidue for any prime p that is congruent to 3 modulo 8. Next let s look at the primes that are congruent to 7 modulo 8, say p = 8k + 7. Now the even numbers 2, 4,..., p 1 are the numbers from 2 to 8k + 6, and the midpoint is 1 2 (p 1) = 4k + 3. The cutoff in this case is (4k + 2) (4k + 4) (4k + 6) (8k + 6). There are exactly 2k + 2 numbers to the right of the vertical bar, so we get 2k + 2 minus signs. This yields 2 (p 1)/2 ( 1) 2k+2 1 (mod p), so Euler s criterion tells us that 2 is a quadratic residue. This proves that 2 is a quadratic residue for any prime p that is congruent to 7 modulo 8. 22

19 [Chap. 24] Is 1 a Square Modulo p? Is 2? 23 Exercises Determine whether each of the following congruences has a solution. (All of the moduli are primes.) (a) x 2 1 (mod 5987) (c) x x 35 0 (mod 337) (b) x (mod 6781) (d) x 2 64x (mod 3011) [Hint. For (c), use the quadratic formula to find out what number you need to take the square root of modulo 337, and similarly for (d).] Use the procedure described in the Primes 1 (Mod 4) Theorem to generate a list of primes congruent to 1 modulo 4, starting with the seed p 1 = Here is a list of the first few primes for which 3 is a quadratic residue and a nonresidue. Quadratic Residue: p = 11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109 Nonresidue: p = 5, 7, 17, 19, 29, 31, 41, 43, 53, 67, 79, 89, 101, 103, 113, 127 Try reducing this list modulo m for various m s until you find a pattern, and make a conjecture explaining which primes have 3 as a quadratic residue Finish the verification of Quadratic Reciprocity Part II for the other two cases: primes congruent to 1 modulo 8 and primes congruent to 5 modulo Use the same ideas we used to verify Quadratic Reciprocity (Part II) to verify the following two assertions. (a) If p is congruent to 1 modulo 5, then 5 is a quadratic residue modulo p. (b) If p is congruent to 2 modulo 5, then 5 is a nonresidue modulo p. [Hint. Reduce the numbers 5, 10, 15,..., 5 2 (p 1) so that they lie in the range from 1 2 (p 1) to 1 2 (p 1) and check how many of them are negative.] Suppose that q is a prime number that is congruent to 1 modulo 4, and suppose that the number p = 2q + 1 is also a prime number. (For example, q could equal 5 and p equal 11.) Show that 2 is a primitive root modulo p. 23

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