Quadratic Reciprocity. As in the previous notes, we consider the Legendre Symbol, defined by

Transcription

1 Math 0 Sring 01 Quadratic Recirocity As in the revious notes we consider the Legendre Sybol defined by $ ˆa & 0 if a 1 if a is a quadratic residue odulo. % 1 if a is a quadratic non residue We also had several roerties of the Legendre sybol: ˆa 1. 1 if and only if x a od q has a solution ˆ ˆa b. If a b od q then ˆa. a 1 od q ˆab ˆa ˆa 1 ˆ b ˆ 1 1q 1 # 1 if 1 od 4q 1 if 1 od 4q. Here we discuss Gauss s law of quadratic recirocity: Theore 1 (Quadratic Recirocity) Let and q be distinct odd ries. Then # ˆ q 1 1 if 1 od 4q 1 if 1 od 4q # ˆ 1 if 1 or 7 od 8q. 1 if or od 8q ˆ. if 1 od 4q or q 1 od 4q 4. q ˆ q if od 4q and q od 4q. Parts 1 and are called the suleentary laws. For coactness we often write art in the for ˆ 1q 1q{8

2 and cobine arts and 4 into the for ˆ q ˆ1 Soe exales. Consider ˆ ˆ ˆ1 4 or using roerty in a tricky way ˆ1 ˆ 11 1q 1 q 1.. We could do either ˆ 1 ˆ ˆ ˆ11 1q 1q Soe ore exales: ˆ ˆ ˆ ˆ0 1q ˆ Here 61 od 8q so 1. One ore: 61 ˆ ˆ ˆ ˆ ˆ61 ˆ11 1q 79 ˆ61 ˆ11 79 ˆ 1q 1 ˆ ˆ ˆ1 ˆ1 1. ˆ ˆ79 1q For a different kind of roble suose we wish to know which ries have the roerty that is a quadratic residue odulo. That is we wish to know which ries have the roerty that x od q have solutions. does not work does (with x 0). ˆ For ą we calculate and note that the squares odulo are 1 and 4. This eans that x od q has a solution for ą if 1 or 4 od q but not if or od q. Thus for exale 16 od q so x od 16q does not have a solution but 79 4 od q so x od 79q has solutions. The Jacobi Sybol There is one final aid to evaluating the Legendre sybol. We introduce a generalization called the Jacobi sybol. With the Jacobi sybol we do not require the denoinator araeter to be rie just an odd nuber larger than 1. More concretely if we write 1 k then the Jacobi sybol is defined by a ˆ ˆ ˆ a a a. 1 k Page

3 Theore (Proerties of the Jacobi sybol) Let and n be odd integers. ˆab a ˆ b 1. a a a. n n a ˆ b. If a b od q then ˆ q 1q{. 1q 1q{8 ˆ 6. If and n are relatively rie then n 1q 1 n 1 n. All of these can be roved by induction on the nuber of ries dividing. If only one rie divides we have the Legendre sybol and this allows the induction to get started. The oint of the Jacobi sybol is that we can send less tie trying to factor ters. For exale using the Jacobi sybol we can reeat soe of our calculations: ˆ ˆ ˆ0 ˆ1 ˆ61 1 1q which was significantly shorter than the calculation with Legendre sybols For another exale using Legendre sybols ˆ ˆ ˆ1 ˆ8 ˆ ˆ ˆ1 ˆ1 ˆ 9 1q 1q Using Jacobi sybols ˆ1 8 ˆ8 1 ˆ161 1 ˆ ˆ Gauss s Criterion How did Gauss rove his law of quadratic recirocity? In fact he liked the theore so uch that over his lifetie he gave soe 7 roofs. I think soe 10 roofs are known today. I will not give a roof but will ention how one of Gauss s roofs went. He started with the following. Page

4 Theore (Gauss s Criterion) Let be an odd rie and let a be an integer not divisible by. For each ositive integer k with k ď 1 let r be the least residue of ka od q. That is r is the unique integer such that 0 ă r ă and r ka od q. Let t be the nuber of r-values for which 1 ˆa ă r ă. Then 1q t. For exale suose 19. I will give a table for calculating the r and t values if a or 4. In each case we need to ultily these nubers by k with 1 ď k ď k t k k k ˆ Fro the table we see that t for a so 1q 1. Siilarly for a ˆ 19 ˆ 4 we have 1q 6 1 and 1q Proof: Let ` 1. Then there are values of k. Let u 1 u... u t be the reainders larger than and let v 1 v... v t be the reainders less than or equal to. For exale when 19 and a fro the iddle row of our table theu s are and the v s are Since k 1 a k a od q only when k 1 k od q which haens only when k 1 k (all k values being less than ) all the reainders are different. This eans all the u s are different and all the v s are different. Now consider the list u 1 u... u t v 1 v... v t. In our exale this would be We see here that we have all the nubers fro 1 to 9. We clai in general that the list is a reordering of the nubers fro 1 to. To verify the clai since the u s are different and the v s are different we need show that u i can t be v j for any i and j. If we had u i v j then u i ` v j ka ` la od q for soe integers k and l. This eans k ` lqa 0 od q. Since ffl a k ` l 0 od q. But k and l coe fro a list of nubers fro 1 to so ď k ` l ă ă so we can t have k ` l 0 od q establishing the clai. Having verified the clai if we ultily the list together then u 1 q u q u t qv 1 v v t u 1 q u q u t qv 1 v v t 1q t u 1 u u t v 1 v v t 1q t r 1 rq r 1q t aqaq aq 1q t a! od q. Page 4

5 But also by the clai u 1 q u q u t qv 1 v v t! od q. Putting these together 1q t a!! od q so 1q t a 1 od q. Since 1 ˆa a od q giving desired. ˆa 1q t as # ˆ 1 if 1 or 7 od 8q Once we have Gauss s Criterion we can show that 1 if or od 8q. We aly Gauss s Criterion with a. The list of nubers k s is and we need t the nuber of these larger than. That is we want the nuber of k so that ă k ď. This nuber is Q ` 1 U t ` 1. We need to know whether t is even or odd. For this we look at od 8q. For exale if 1 od 8q then 8n ` 1 for soe n and 4n. In this case Q 4n ` 1 U t 4n ` 1 4n n ` 1q ` 1 n an even nuber. In general writing 8n ` r 4n ` r 1 with r 1 7 we have t 4n ` r 1 Q 4n ` r 1 ` 1U 4n ` 1 ` r 1 Q r ` 1 n n ` r ` 1 Q r ` 1 4 and lugging in r 1 7 gives t n n ` 1 n ` 1 n ` and we get () of Quadratic Recirocity. I will not rove () and (4) of Quadratic Recirocity but you can now guess how it ˆ ight go: We use Gauss s Criterion with 1q t 1 and 1q t. This eans q ˆ 1q t 1`t. Much clever reasoning later one gets 1q t 1`t 1q 1 q 1. See q the book ages or the internet for the details. U 4 ` 1 U Page

6 Soe sale robles and solutions. Here are soe robles to ractice on for the final exa. 1. Evaluate each of the following Legendre sybols. ˆ ˆ ˆ ˆ 6 10 ˆ Evaluate the last one with/without the Jacobi sybol.. What can you say about a rie if n `?. What can you say about a rie if n `? Use this to show that there are infinitely any ries of the for 6k ` 1. Solutions 1. Evaluate each of the following Legendre sybols. ˆ (a) (b) Solution: Since 1 od 4q inverting does not change a sign: ˆ ˆ ˆ 1. ˆ I don t use the rule for here since is sall we can just figure out what the residues and nonresidues are. For 1 is a residue and is a nonresidue. Siilarly for 1 and 4 are residues and are nonresidues. I use this aroach for sall ries. ˆ7 9 Solution: ˆ ˆ ˆ7 ˆ since and 9 are both congruent to odulo 4 quadratic recirocity says a negative sign is introduced when inverting. Continuing ˆ9 ˆ 1. Page 6

7 (c) ˆ Solution: ˆ ˆ641 1 ˆ1 1 aking use of Jacobi sybol roerties and ˆ ˆ ˆ1 ˆ ˆ1 1 ˆ 1 since 1 od 8q. Finishing it off ˆ ˆ1 ˆ (d) ˆ 6 10 Solution: ˆ ˆ ˆ ˆ ˆ since 1 due to the fact that 10 7 od 8q 10 ˆ ˆ ˆ ˆ 1 ˆ10 ˆ ˆ1 ˆ1 1. (e) ˆ71 both with/without Jacobi sybols. 919 Solution: Without: ˆ71 ˆ919 ˆ ˆ ˆ71 ˆ71 1q 1q 71 7 With Jacobi Sybols goes the sae till: ˆ ˆ ˆ71 ˆ ˆ 71 ˆ1 This last ste was because 16 is a erfect square. ˆ 71 ˆ q1q 1. ˆ 7 ˆ71 ˆ Page 7

8 . What can you say about a rie if n `? Solution: First if n is even can be. For odd ries n ` is equivalent to n od q and this is ossible only if 1. Now ˆ ˆ ˆ 1 ˆ. If we work with od 8q the cases are 1 7 leading to 1q1q 1q 1q 1q 1q 1q1q giving a sign attern of That is the ossible odd ries dividing n ` are this congruent to 1 or odulo 8.. What can you say about a rie if n `? Use this to show that there are infinitely any ries of the for 6k ` 1. Solution: If n is odd could be and if n is divisible by then could be. For ries larger than as with the revious roble this is equivalent to n ˆ od q having a solution so we need those ries with 1. Here ˆ ˆ 1 ˆ 1q 1q{ 1q 1 1 1q 1 1 1q. This eans we need 1 od q since 1 is the only quadratic residue for. Since also has to be even the ossible ries dividing n ` are and ries 1 od 6q. Now we rove that there are infinitely any ries of the for 6k`1. Certainly there are ries of this for: etc. Suose there were only finitely any say 1... k. We now use the first art in a tricky way: Let M 4 1 k q `. This nuber is not divisible by or by. If q is a rie divisor of this nuber then q n ` where n 1 k so q 1 od 6q. Also q can t be one of the s because i M says i and none of these ries divides. Thus our list was not colete a contradiction. Page 8