Justin Cantu Math 482 Texas A&M University Instructor: Dr. David Larson Parts I and II of the Law of Quadratic Reciprocity

Quadratic Residues and Nonresidues

The Law of Quadratic Reciprocity tells us whether or not the congruence x2 ≡ a (mod p) has a solution. In other words, whether a is a square modulo p. Example: Does x2 ≡ 3 (mod 7) have a solution? We can square the numbers 0 to 6, reduce modulo 7, and see if any of them is equal to 3: 02 ≡ 0 (mod 7) 12 ≡ 1 (mod 7) 22 ≡ 4 (mod 7) 32 = 9 ≡ 2 (mod 7) 42 = 16 ≡ 2 (mod 7) 52 = 25 ≡ 4 (mod 7) 62 = 36 ≡ 1 (mod 7) So x2 ≡ 3 (mod 7) does not have a solution. To look for patterns and make conjectures, we create a table of all squares modulo p for a few primes. This is a usual procedure in Number Theory.

Modulo 5 Modulo 7 Modulo 11 Modulo 13 2 b b2 b b2 b b2 b b

0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1

2 4 2 4 2 4 2 4

3 4 3 2 3 9 3 9

4 1 4 2 4 5 4 3

5 4 5 3 5 12

6 1 6 3 6 10

7 5 7 10

8 9 8 12

9 4 9 3

10 1 10 9 11 4

12 1 We can see that each number that appears as a square seems to appear exactly twice. Also, there is actually a sort of reverse repeating pattern in each table. What this pattern amounts to is saying b2 and (p-b)2 are the same modulo p. This is easy to see since: (p-b)2 = p2 -2pb + b2 ≡ b2 (mod p).

Because of this pattern, if we want to list all the nonzero numbers that are squares modulo p, we only need to compute half of them, 12 (mod p), 22 (mod p), …, ((p-1) / 2)2 (mod p).

Our goal is to find patterns that can be used to distinguish squares from nonsquares modulo p.

Definition: A nonzero number that is congruent to a square modulo p is called a quadratic residue modulo p and is abbreviated QR. Looking at the tables, we see 3 and 12 are QRs modulo 13.

Definition: A number that is not congruent to a square modulo p is called a (quadratic) nonresidue modulo p and is abbreviated NR. For example, 2 and 5 are NRs modulo 13.

A number that is congruent to 0 modulo p is neither a residue nor a nonresidue.

Listing the QRs and NRs modulo 13, we have QRs mod 13 : {1, 3, 4, 9, 10, 12} NRs mod 13: {2, 5, 6, 7, 8, 11} and we notice there are 6 of each.

Listing the QRs and NRs modulo 7, we have QRs mod 7: {1, 2, 4} NRs mod 7: {3, 5, 6} and we notice there are 3 of each. This leads to the following theorem.

Theorem 1: Let p be an odd prime. Then there are exactly (p-1)/2 quadratic residues modulo p and exactly (p-1)/2 nonresidues modulo p.

Proof: The quadratic residues are the nonzero numbers that are squares modulo p, so they are the numbers 12 , 22, …, (p-1)2 (mod p). But we really only need to go halfway since b2 ≡ (p-b)2: 12 , 22, …, ((p-1)/2)2 (mod p). Since the list is of size (p-1)/2, we need to check that these numbers are all different modulo p. 2 2 Suppose that b1 and b2 are numbers between 1 and (p-1)/2 and that b1 ≡ b2 (mod p). We want to 2 2 2 2 show b1 = b2. The fact that b1 ≡ b2 (mod p) means that p divides b1 - b2 = (b1-b2)(b1+b2). Since b1+b2 is between 2 and p-1, p must divide b1-b2. But |b1-b2| < (p-1)/2, so the only way for b1-b2 to 2 2 2 be divisible by p is to have b1-b2 = 0 or b1 = b2. Thus 1 , 2 , …, ((p-1)/2) are all different modulo p and there are (p-1)/2 quadratic residues. Clearly there are p-1 numbers between 1 and p-1. Since (p-1)/2 are QRs, the other (p-1)/2 must be NRs.

What happens when we multiply a QR × QR, QR × NR, or NR × NR? For example, 3 and 10 are QRs modulo 13 and 3•10 = 30 ≡ 4 (mod 13), a QR. This case is easy 2 2 to see. Suppose a1 and a2 are both QRs modulo p, then a1 ≡ b1 (mod p) and a2 ≡ b2 (mod p) for 2 2 2 some numbers b1 and b2. So a1a2 ≡ b1 b2 ≡ (b1b2) (mod p), or a1a2 is a QR. What about QR × NR and NR × NR? Again, we can try a few values:

QR × NR ≡ ?? (mod p) NR × NR ≡ ?? (mod p)

2 × 5 ≡ 3 (mod 7) NR 3 × 5 ≡ 1 (mod 7) QR

5 × 6 ≡ 8 (mod 11) NR 6 × 7 ≡ 9 (mod 11) QR

4 × 5 ≡ 7 (mod 13) NR 5 × 11 ≡ 3 (mod 13) QR

10 × 7 ≡ 5 (mod 13) NR 7 × 11 ≡ 12 (mod 13) QR

It seems that QR × NR = NR and NR × NR = QR. This leads to our next theorem.

Theorem (Quadratic Residue Multiplication Rule): Let p be an odd prime. Then: i) QR × QR = QR ii) QR × NR = NR iii) NR × NR = QR.

2 Proof: We already proved (i). Suppose that a1 is a QR, so a1 ≡ b1 (mod p), and a2 is an NR. 2 Assume that a1a2 is a QR to derive a contradiction. So a1a2 ≡ b3 for some b3 and we have 2 2 b3 ≡ a1a2 ≡b1 a2 (mod p). 2 Since p doesn’t divide a1 = b1 , p doesn’t divide b1, thus gcd(b1, p) = 1. So b1 has an inverse 2 modulo p, say ∃c1 such that c1b1 ≡ 1 (mod p). Multiplying the above by c1 we get, 2 2 2 2 c1 b3 ≡ c1 a1a2 ≡ (c1b1) a2 ≡ a2 (mod p). 2 Thus a2 ≡ (c1b3) (mod p) is a QR, a contradiction. We will need the following lemma to prove (iii).

Lemma: Let p be a prime number and let a be a number with a ≠ 0 (mod p). Then the numbers a, 2a, 3a, …, (p-1)a (mod p) are the same as the numbers 1, 2, 3, …, (p-1) (mod p), although they may be in a different order.

Proof of (iii): Let a be an NR and consider the set of values a, 2a, 3a, …, (p-2)a, (p-1)a (mod p). By the lemma, these are just the numbers 1, 2, …, (p-1) rearranged in some different order. In particular, they include the (p-1)/2 QRs and the (p-1)/2 NRs. However, as we already proved, each time that we multiply a by a QR, we get an NR, so the (p-1)/2 products a × QR already give us all (p-1)/2 NRs in the list. Hence when we multiply a by an NR, the only possibility is that it is equal to one of the QRs in the list, because the a × QR products have already used up all of the NRs in the list.

Looking at the three equations i) QR × QR = QR ii) QR × NR = NR iii) NR × NR = QR we see that QR behaves like 1 and NR behaves like -1. Adrien-Marie Legendre introduced the Legendre symbol of a modulo p as ! a $ = 1 if a is a quadratic residue modulo p, "# p%& { -1 if a is a nonresidue modulo p.

For example, looking at the table on page 1, we see (3/13) = 1 (11/13) = -1 (2/7) = 1 (3/7) = -1.

Theorem 2: Let p be an odd prime. Then (a/p)(b/p) = (ab/p).

Example: (75/97) = (3•5•5/97) = (3/97)(5/97)(5/97) = (3/97)(±1)2 = (3/97). Since 102 = 100 ≡ 3 (mod 97), 3 is a QR modulo 97. Thus (75/97) = (3/97) = 1.

Law of Quadratic Reciprocity Part I

We took a prime p and found the QRs and NRs. Now we are going to fix a and find which primes p have a as a QR.

First we set a = -1 and ask for which primes p is -1 a QR? In other words, for which primes p does the congruence x2 ≡ -1 (mod p) have a solution? Again, we use a table to motivate us.

p 3 5 7 11 13 17 19 23 29 31

Solutions to NR 2, 3 NR NR 5, 8 4, 13 NR NR 12, 17 NR x2 ≡ -1 (mod p)

Conjecture: " ! 1 % 1 if p ≡ 1 (mod 4) = #$ p &' { -1 if p ≡ 3 (mod 4) We first prove the Square Root of Fermat’s Little Theorem and Euler’s Criterion to help prove our conjecture.

Square Root of Fermat’s Little Theorem: Let A = a(p-1)/2, then A ≡ ±1 (mod p).

Proof: By Fermat’s Little Theorem, A2 = ap-1 ≡ 1 (mod p). Thus p divides A2-1 = (A-1)(A+1), so either p divides A-1 or A+1. Thus A ≡ 1 (mod p) or A ≡ -1 (mod p).

Let us make a table with some values of p, a, A (mod p), and (a/p).

p 11 31 47 97 173 409 499 601

a 3 7 10 15 33 78 33 57

A (mod p) 1 1 -1 -1 1 -1 1 -1

(a/p) 1 1 -1 -1 1 -1 1 -1 It seems that A ≡ 1 when a is QR and A ≡ -1 when a is NR.

Euler’s Criterion: Let p be an odd prime. Then A = a(p-1)/2 ≡ (a/p) (mod p).

Proof: Suppose that a is a quadratic residue, say a ≡ b2 (mod p). By Fermat’s Little Theorem, we have a(p-1)/2 ≡ (b2)(p-1)/2 ≡ bp-1 ≡ 1 ≡ (a/p) (mod p). Next, consider the congruence X(p-1)/2 - 1 ≡ 0 (mod p). We just proved that every quadratic residue is a solution to this congruence and we know from earlier that there are exactly (p-1)/2 quadratic residues. Also, this congruence has at most (p-1)/2 solutions. Thus, the solutions to this congruence are exactly the QRs modulo p.

Now let a be a nonresidue. By Fermat’s Little Theorem, a(p-1) ≡ 1 (mod p), or 0 ≡ ap-1 - 1 ≡ (a(p-1)/2 - 1)(a(p-1)/2 + 1) (mod p). The second factor must be 0, or a(p-1)/2 + 1 ≡ 0, which implies a(p-1)/2 ≡ -1 = (a/p) (mod p).

Example: Is -1 a quadratic residue modulo p = 6911? (-1)(6911-1)/2 = (-1)3455 = -1. Euler’s Criterion tells us that (-1/6911) ≡ -1 (mod 6911). Since (a/p) is either 1 or -1, we have (-1/6911) = -1, thus -1 is a nonresidue modulo 6911.

Theorem (Quadratic Reciprocity Part 1): Let p be an odd prime. Then " !1% 1 if p ≡ 1 (mod 4) ={ #$ p &' -1 if p ≡ 3 (mod 4).

Proof: Suppose that p ≡ 1 (mod 4), say p = 4k + 1. Then (-1)(p-1)/2 = (-1)2k ≡ 1, so 1 ≡ (-1/p) (mod p), but actually 1 = (-1/p).

Suppose that p ≡ 3 (mod 4), say p = 4k + 3. Then (-1)(p-1)/2 = (-1)2k+1 ≡ -1, so -1 ≡ (-1/p) (mod p), but actually -1 = (-1/p).

Law of Quadratic Reciprocity Part II

Now we set a = 2, and ask, for which primes p is 2 a QR? 2 is a QR for p = 7, 17, 23, 31, 41, 47, 71, 73, 79, 89, 97, 103, 113, 127... 2 is an NR for p = 3, 5, 11, 13, 19, 29, 37, 43, 53, 59, 61, 67, 83, 101, 107, 109...

Reducing these lists modulo 4 gives: 3, 1, 3, 3, 1, 3, 3, 1, 3, 1, 1, 3, 1, 3 3, 1, 3, 1, 3, 1, 1, 3, 1, 3, 1, 3, 3, 1, 3, 1. There is no immediate pattern. Reducing these lists modulo 8 gives: 7, 1, 7, 7, 1, 7, 7, 1, 7, 1, 1, 7, 1, 7 3, 5, 3, 5, 3, 5, 5, 3, 5, 3, 5, 3, 3, 5, 3, 5.

Conjecture: ! 2 $ 1 if p ≡ 1 or 7 (mod 8) = "# p %& { -1 if p ≡ 3 or 5 (mod 8).

We can’t use Euler’s criterion in an obvious way since 2(p-1)/2 (mod p) may be hard to calculate. We use a trick. We begin with an example. Let p = 13 and multiply the first (p-1)/2 = 6 numbers by 2 and then multiply those together. So, 2•4•6•8•10•12 = 26•1•2•3•4•5•6 = 26•6! We next take 2, 4, 6, 8, 10, 12 and reduce them modulo 13 so that they are in the range -6 to 6. 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)

This gives 2•4•6•8•10•12 ≡ 2•4•6•(-5)•(-3)•(-1) ≡ (-1)3•6! ≡ -6! (mod 13) Equating these two values gives 26•6! ≡ -6! (mod 13) which implies 26 ≡ -1 (mod 13).

General Case: Let p be an odd prime and let P = (p-1)/2. We start with the even numbers 2, 4, 6, …, p-1. Multiplying them together and factoring out 2 gives, 2•4•6… (p-1) = 2(p-1)/2•1•2•3… (p-1)/2 = 2P•P!. Now we reduce these same even numbers modulo p so that they are in the range -P to P. The first few numbers won’t change but at some point in the list we’ll start hitting numbers that are larger than P, and each of these large numbers need 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 minus signs = {Number of integers in the list 2, 4, 6, …, (p-1) that are larger than P=(p-1)/2}

Illustration that helps: 2•4•6•8 ••• | ••• (p-5)•(p-3)•(p-1) Numbers ≤ (p-1)/2 are left unchanged | Number > (p-1)/2 need to subtract p from each.

Comparing the two products, we have 2P•P! = 2•4•6•••(p-1) ≡ (-1)(number of minus signs) •P! (mod p) which implies 2(p-1)/2 ≡ (-1)(number of minus signs) (mod p)

Quadratic Reciprocity Part II: Let p be an odd prime. Then ! 2 $ 1 if p ≡ 1 or 7 (mod 8) = "# p%& { -1 if p ≡ 3 or 5 (mod 8).

Proof: Suppose p ≡ 3 (mod 8), say p = 8k+3. We need to find the number of minus signs where p-1 = 8k+2 and (p-1)/2 = 4k+1. We use our illustration to help 2•4•6•••4k | (4k+2)•(4k+4)•••(8k+2). Dividing the right side by 2 gives the numbers (2k+1), (2k+2), …, (4k+1) which is (4k+1)-(2k+1)+1 = 2k+1 numbers, thus 2k+1 minus signs. So 2(p-1)/2 ≡ (-1)2k+1 ≡ -1 (mod p). By Euler’s Criterion, 2 is a nonresidue modulo p.

Suppose p ≡ 7 (mod 8), say p = 8k+7. We need to find the number of minus signs where p-1 = 8k+6 and (p-1)/2 = 4k+3. We use our illustration to help 2•4•6•••(4k+2) | (4k+4)•(4k+6)•••(8k+6). This gives (4k+3)-(2k+2)+1 = 2k+2 minus signs. So 2(p-1)/2 ≡ (-1)2k+2 ≡ 1 (mod p). By Euler’s Criterion, 2 is a quadratic residue modulo p. The last two cases are similar. References

Silverman, Joseph H. A Friendly Introduction to Number Theory. 4th Ed. Upper Saddle River: Pearson, 2013. 141-157. Print.