Quadratic Reciprocity

Transcription

1 Quadratic Recirocity Quadratic recirocity relates solutions to x = (mod to solutions to x = (mod, where and are distinct odd rimes. The euations are oth solvale or oth unsolvale if either or has the form 4k + 1; one is solvale and one is unsolvale if oth rimes have the form 4k + 3. Quadratic recirocity can e exressed in terms of Legendre symols: If and are distinct odd rimes, then = if either or has the form 4k + 1, whereas = if oth rimes have the form 4k + 3. Lemma. (Gauss Let e an odd rime, (a, = 1. Let k e the numer of least ositive residues of a, a,..., 1 a that are greater than. Then a = ( 1 k. Proof. Since is odd, 1 is not an integer. Hence, every residue of a, a,..., a is either less than or greater than. Lael these two sets: Thus, j + k = 1. Ste 1 1,..., k, a 1,..., a j } = The a i s are contained in 1. What aout the i s? a 1,..., a j <, 1,..., k >. 1,,..., 1 }. 1,,..., 1 }, ecause the a i s are less than i >, so i < =. (so less than or eual to Since i is an integer and is an integer lus one-half, I have i 1. This shows that the i s are contained in 1,,..., 1 } as well. There are 1 elements in 1,,..., 1 }, and j + k = 1. So if the a i s and i s are all distinct, I ll know the two sets are eual. Each a i has the form ra, where 1 r 1. So if ra and sa are the same, then ra = sa (mod, so (r sa. a, so (r s. This is imossile for 1 r, s 1 unless r = s which imlies ra = sa to egin with. 1

2 A similar argument shows that the i s, and hence the i s, are distinct. Could a i = h? a i = ra and h = sa for 1 r, s 1, so sa = ra (mod, ra + sa = 0 (mod, (r + sa. Again, a, so (r + s. But 1 r, s 1 This finishes the roof that 1,..., k, a 1,...,a j } = imlies r + s 1, so (r + s is imossile. 1,,..., 1 }. Ste Since the two sets are the same, the roducts of the elements in the two sets are the same: ( 1 ( 1 ( k a 1 a j =! (mod. Now i = i (mod, so ( 1 ( 1 k 1 k a 1 a j =! (mod. But the a s and s are exactly the residues of the numers a, a,..., 1 a, so I may relace the roduct of the a s and s with the roduct of a, a,..., 1 a: ( ( 1 1 ( Now, 1 theorem, I get ( 1 k a a a = ( 1 ( 1 k a ( 1/! = ( 1 = 1, so I can cancel the ( 1 k a ( 1/ = 1 (mod, ( 1 ( 1 k a = 1 (mod,! (mod,! (mod.! terms from oth sides. Then alying Euler s a = ( 1 k (mod. I made the last ste y multilying oth sides y ( 1 k and using the fact that ( 1 k = 1. Examle. I ll use Gauss s lemma to comute 6 7 Since = 7, 1 = 3.5 and = 3. Look at the residues 1 6 = 6, 6 = 5, and 3 6 = 4. All three are greater than 3.5 they re i s, in the notation of the roof of Gauss s lemma so Gauss says 6 = ( 1 3 = 1. 7 As a check, Euler s theorem gives 6 = 6 3 = 1 (mod 7. 7 Lemma. Let a, > 0, where is an odd integer. Then ([ a ] a = + e + ( 1 e r,

3 where e = 0 or 1 and 0 r 1. Here denotes the greatest integer function and Proof. By the Division Algorithm, a = + r, where 0 r <. [ a ] + e is the integer closest to a. Now is not an integer, so either r < or r >. (For examle, if a = 11 and = 3, then r = > 3 =, while if a = 11 and = 5, r = 1 < 5 =. Consider the two cases. Case 1: r <. Write ([ a ] a = ( 1 0 r. Here e = 0, and 0 r 1. Case : r >. [ a ] + 0 is the integer closest to a. r <, ut is not an integer, so r 1, and Write ([ a ] ([ a ] a = (r = ( 1 1 ( r. [ a ] Here e = 1, and + 1 is the integer closest to a. r <, so r > 0. r >, so r <, or r < =. Since is not an integer, r 1. Therefore, 0 r 1. Examle. Take a = 4 and = , so the integer closest to is = , and Take a = 50 and = =.94, so the integer closest to is = (

4 In other words, the r in the lemma reresents the distance from a to the nearest multile of. In the first case, the nearest multile is to the left... r= while in this case, the nearest multile is to the right. r= The ± is needed deending on whether the nearest multile is less than or greater than a. I ll use the lemma to give an ingenious roof of Quadratic Recirocity due to J.S. Frame [1]. Theorem. (Quadratic Recirocity Let and e distinct odd rimes. 1 1 = ( 1 Proof. To simlify the writing, let = 1 and = 1. Let 1 n. Aly the Lemma with a = and = : = ( + e n + ( 1 en r n, where 1 r n and e n = 0 or 1. The first thing I will show is that the remainders r n are just a ermutation of the integers 1,...,. If I take the initial euation and go mod, I get = ( 1 en r n (mod. Can two of the r s e eual? Suose r m = r n, where 1 m, n. Then 0 = r m r n = (( 1 em m ( 1 en n (mod. In other words, (( 1 em m ( 1 en n. But m + n = 1, so ( 1 em m ( 1 en n is surely smaller than in asolute value. Since, this is imossile unless ( 1 em m ( 1 en n = 0. This in turn is imossile unless m = n. Thus, the r s are distinct. Since there are of them, and since they re all in the range [1, ], they must e some ermutation of the numers 1,...,. As a reliminary to the next comutation, take the first euation and go mod. and are odd, so they eual 1 mod ; ( 1 en = ±1, so either way it euals 1 mod. Therefore, n = + e n + r n (mod. 4

5 (I m going to use this in an exonent of 1 in a second! Now let 1 m, 1 n. Then m 0, for m = imlies m imossile, ecause 1 m = 1. Now here s the heart of the roof. The idea will e to define a weird roduct which turns out to e the Legendre symol. Define f(, = m=1 m m. m Notice that is a fancy way of exressing the sign of m +1 when it s ositive, 1 m when it s negative. When is m negative? m < gives m <, or m. That is, m is negative for m = 1,...,. So the roduct (for fixed n has 1 s, and f(, = ( 1 [/]. Now n = + e n + r n (mod, so n r n e n = (mod. In fact, e n = e n (mod, so [ ] n r n + e n = (mod. Since things which are eual mod give the same ower of 1, I have f(, = ( 1 n rn+en = ( 1 n rn ( 1 en = ( 1 (n rn ( 1 en. Since the r n s are just the integers from 1 to and since n runs from 1 to, (n r n = 0! So now f(, = ( 1 en. If I take the very first euation and go mod, I get = ( 1 en r n (mod, or r n = ( 1 en (mod. (r n is invertile mod, so the fraction makes sense. So now f(, = r n (mod. But n r n = 1, ecause as n runs over the numers from 1 to, so does r n. So y Euler s theorem, f(, = = = 5

6 Notice that f(, = m=1 m n m n = m=1 I got the second roduct y swaing m and n in the first. Whew! The rest is easy fortunately! = f(, f(, = Since = 1 m=1 and = 1, I m done! m m. m m m m = m=1 ( 1 = ( 1. As comlicated as this roof is, it s actually no worse than most roofs of this result. Before giving an examle, I want to discuss what recirocity tells you aout solutions to uadratic congruences. An odd rime is congruent to 1 or to 3 mod 4. If = 4k + 1, then 1 = k, an even numer. If = 4k + 3, then 1 = k + 1, an odd numer. Since an even numer times anything is even, 1 1 = even if or = 1 (mod 4 odd if and = 3 (mod 4 Therefore, However, 1 1 = ( 1 +1 if or = 1 (mod 4 = 1 if and = 3 (mod 4 = +1 means = = 1 or = = 1 = 1 means one of, is + 1 and the other is 1 In terms of the congruences x = (mod and x = (mod this means: 1. If at least one of, is congruent to 1 mod 4, then oth euations are solvale or oth euations are unsolvale.. If oth and are congruent to 3 mod 4, then one euation is solvale and the other is unsolvale. Corollary. Let and e distinct odd rimes. (a If at least one of, is congruent to 1 mod 4, then =. ( If oth and are congruent to 3 mod 4, then =. 6

7 Examle. Comute = 1 (mod 4, so 17 = 71 = 3 = 17 = = (3 1/ = = 1 (mod In other words, x = 17 (mod 71 does not have any solutions. Examle. Comute = 13 3 ; I ll comute 13 and = 359 = 8 = 8 (13 1/ = 8 6 = 6144 = 1 (mod 13, = 359 = 14 = Next, I ll comute and = (3 1/ = 11 = 048 = 1 (mod 3, 3 7 = 3 = = ( (7 1/ = 8 = 1 (mod Therefore, 3 = (1( 1 = 1, and = ( 1(1 = The congruence x = 99 (mod 359 does not have a solution. [1] J.S. Frame, A short roof of uadratic recirocity, Amer. Math. Monthly, 85(10(1978, c 011 y Bruce Ikenaga 7