Lecture 9 Quadratic Residues, Quadratic Recirocity Quadratic Congruence - Consider congruence ax + bx + c 0 mod, with a 0 mod. This can be reduced to x + ax + b 0, if we assume that is odd ( is trivial case). We can now comlete the square to get ( a ) a x + + b 0 4 mod So we may as well start with x a mod If a 0 mod, then x 0 is the only lution. Otherwise, there are either no lutions, or exactly two lutions (if b a mod, then x ± b mod ). (x a b mod x b (x b)(x + b) x b or b mod ). We want to know when there are 0 or lutions. (Definition) Quadratic Residue: Let be an odd rime, a 0 mod. We say that a is a quadratic residue mod if a is a square mod (it is a quadratic non-residue otherwise). Lemma 39. Let a mod 0 mod. Then a is a quadratic residuemod iff a Proof. By FLT, a mod and is even. This follows from index calculus. Alternatively, let s see it directly ( a ) mod a ± mod Let g be a rimitive root mod., g, g... g },,... } mod. Then a g k mod for me k. With that a g k+( )m mod k s only defined mod. In articular, since is even, we know k is even or odd doesn t deend on whether we shift by a multile of. (ie., k is well defined mod ). We know that a is quadratic residue mod iff k is even (if k l then a g l (g l ) mod ). Conversely if a b mod and b g l mod we get a g l mod, k is even. Note: this shows that half of residue class mod are quadratic residues, and half are quadratic nonresidues. Now look at ( ) k( ) a g k g mod. k mod iff ord g divides k( ) iff ( ) k( ) k a is a quadratic residue.
(Definition) Legendre Symbol: ( a ) if a is a quadratic residue mod if a is a quadratic non-residue mod Defined for odd rime, when (a, ). (For convenience and clarity, written (a )). We just showed that (a ) a mod. Remark. This formula shows us that (a )(b ) (ab ). LHS a b (ab) mod RHS mod and since both sides are ± mod, which is an odd rime, they must be equal Similarly, (a ) (a ) Eg. ( 4 79) ( 79) ( 79)( 79) ( 79) ( ) 39 Al, 79 is not mod 4 is quadratic non-residue. We ll work toward quadratic recirocity relating ( q) to (q ). We ll do Gauss s 3rd roof. Lemma 40 (Gauss Lemma). Let be an odd rime, and a 0 mod. For any integer x, let x be the residue of x mod which has the smallest ablute value. (Divide x by, get me remainder 0 b <. If b >, let x b, if b >, let x be b. ie., < x < ) Let n be the number of integers among (a), (a), (3a)... (( )a) which are negative. Then (a ) ( ) n. Proof. (Similar to roof of Fermat s little Theorem) We claim first that if k l then (ka) ±(la). Suose not true: (ka) ±(la). Then, we d have ka ±la mod (k l)a 0 mod k l 0 mod This is imossible because k + l and < k l < and k l 0 (no multile of ossible). So the numbers (ka) for k... are all distinct mod (there s of
them) and must be the integers, 3... } in me order. ( ) (ka) mod ( ) n (ka) mod ( ) n k ka mod ( ( a )) ( ) n mod a ( ) n mod a ( ) n mod (a ) ( ) n mod (a ) ( ) n since > where the second ste follows from the fact that exactly n of the numbers (ka) are < 0. Theorem 4. If is an odd rime, and (a, ), then if a is odd, we have (a b) ( ( ) t )/ ja ( where t )/. Al, ( ) ( ) j Proof. We ll use the Gauss Lemma. Note that we re only interested in ( ) n. We only care about n mod. We have, for every k between and ka 0 if (ka) > 0 ka + (ka) + if (ka) < 0 ka 0 if (ka) > 0 + (ka) + mod if (ka) < 0 Sum all of these congruences mod 3
( )/ ( )/ ka ka ( )/ + (ka) + n mod ( )/ ( ka a )/ k ( ) ( ) a + a( ) Now (a). Since a,..., a } is just... }, Plug in to get ( )/ ( )/ ( ka) k ( ) ( ) ( ) ( ) ( )/ ka n a + mod ( (a ) / ) ( ) + ( ka ) mod If a is odd, we have is integer and a is even, roduct 0 mod, to get n ( )/ t mod ka So (a ) ( ) n ( ) t mod When a, n + ( )/ k mod 4
So, note that for k... } k k 0 < < k 0 n ( )/ (k ) 0 mod and ( ) ( n ) ( ) So far, if mod 4 ( ) ( ) if 3 mod 4 Check if, 7 mod ( ) ( ) if 3, 5 mod 4 Theorem 4 (Quadratic Recirocity Law). If, q are distinct odd rimes, then q if or q mod 4 ( q)(q ) ( ) otherwise Proof. Consider the right angled triangle with vertices (0, 0), (, 0), (, q ). Note that: no integer oints on vertical side, no nonzero integer oints on hyotenuse (sloe is q, if we had integer oint (a, b) then b q a b qa, a, q b, and if (a, b) (0, 0), then a, b q). Ignore the ones on horizontal side. Claim: the number of integer oints on interior of triangle is ( )/ qk 5
Proof. If we have a oint (k, l), then k and sloe l k l < qk. Number of oints on the segment x k is the number of ossible l, which is qk just. < q Add these (take triangle, rotate, add to make rectangle) - adding oints in interior of rectangle is ( )/ ( l + q )/ l ( ) ( ) qk q (q ) ( ) t where t qk l ( q) ( ) t where t q ( q)(q ) ( ) t+t where t + t total number of oints 6
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