Legendre symbol Correlation

Small watermelon recently learned this thing, do not know very well, want to hit a log to consolidate a bit ...
Legendre symbol Well, it's interesting (this sentence is nonsense), it is related to the sum of squares : Set A, A, is two non-0 integers,b is a prime number, we define the symbol : If there is an integer x, make , then remember; otherwise, remember . When p|a ,=0. (This symbol is a bit troublesome to knock, for convenience, the following (a\b) , the direction of the opposite of Division sign)

Some properties:

First of all, by Euler discriminant conditions, there is obviously (a\b)≡a^ ((b-1)/2) (mod b)

If A1≡a2 then (a1\p) = (a2\p), it is clear that the set of Euler criteria is OK.
Obviously...... ( -1\p) = ( -1) ^ ((p-1)/2)

　　Then, as if I could not think of any nature ...

So get to the point and the next thing is ... Gaussian lemma:
If P is a prime number,(a,p) =1, the record μ is a series of a,2a,3a ... , [(p-1)/2]a the minimum positive remaining is greater than (p-1)/2 A few numbers, then (a\p) = ( -1) ^ μ, Proof ( feeling is a bit like Fermat theorem proving method ) :

withR1,R2,R3...RKindicates that the smallest positive remaining in the sequence is not less than(p-1)/2the number,S1,S2,S3...sμ is the remaining number, which proves{R},P-{s}(To{s}for each number in thePThe numbers in the two seriesP22 Non-reciprocal:
For a number in the same sequence, the conclusion is obvious (see Fermat theorem proof). SetRI≡P-SJ (mod p), youp| (RI+SJ), rememberRi=ua,Si=va, youp| ((V+u) a), since(a,p) =1, sop| (v+u), butv+u<p, it is impossible, so the proof.

So obviously, the sequence{R},P-{s}will become,..., (p-1)/2of an arrangement, then we put{R},{s}to connect and get a ride .R1*R2*R3...RK*S1*S2*S3...sμ≡( -1) ^Μ((p-1)/2)! (mod p), i.e.a*2a*3a*...((p-1)/2) a≡( -1) ^Μ* ((p-1)/2)! (mod p), soa^ ((p-1)/2)≡( -1) ^Μ(mod p), so(a\p) = ( -1) ^μ.

Finally finished ...
The Gaussian lemma is a magical thing that can often be used when doing a problem.

It's the climax! And then there are two reciprocal laws (and if I know it all, it's a bluff):

Set p,q is two different singular primes, then (p\q) (q\p) = ( -1) ^ ((p-1)/2+ (q-1)/2), the front content with the Gaussian lemma proof, We go on to discuss:

     Rememberr=ΣR,s=Σs, then σr+Σ(p-s) =r+μp-s, as had been said at the time of the proof of the Gaussian lemma,R1,R2,R3... ..rk,p-s1,p-s2,... p-sμ is,.., (p-1)/2of an arrangement, then(A ) * ((P-1)/2) * ((p-1)/2+1) = (P²-1)/8=r+μp-s, and then it's easy to pushr=s-μp+ (P²-1)/8.

Next, consider all the formulas:Iq=[iq/p]p+ti (with remainder Division), if you put these formulas onISum,Q∑i=pΣ[iq/p]+∑ri+∑si, namelyQ (P²-1)/8=pΣ[Iq/p]+r+sand Willr=s-μp+ (P²-1)/8generation, you have toQ (P²-1)/8=pΣ[iq/p]+2s-μp+ (P²-1)/8, i.e.(q-1) (P²-1)/8=p (Σ[iq/p]-μ) +2s.

because(q-1) (P²-1)/8is an even number (q-1is an even number,(P²-1)/8is an integer)2Sis even, so σ[iq/p]-μ must be even, thus σ[iq/p]-μ can be:( -1) ^Σ[iq/p]= ( -1) ^μ= (q\p), sinceP,Qequivalent, then( -1) ^Σ[ip/q]= ( -1) ^μ= (p\q), multiplied to get( p\q) (q\p) = ( -1) ^Σ[iq/p]* ( -1) ^Σ[ip/q].

The following proves Σ[jp/q]+Σ[iq/p]= (p-1)/2+ (q-1)/2). Take((p-1)/2) * ((Q-1)/2)number,Jp-iq, apparently none of them0(disprove), then consider the positive number, for each givenJto meetJp>iqof theITotal[jp/q]A, andJless than or equal(q-1)/2, so the number of positive numbers is σ[jp/q], in the same vein, the number of negative numbers is σ[iq/p], so I have to pass σ[jp/q]+Σ[iq/p]= (p-1)/2+ (q-1)/2), thus(p\q) (q\p) = ( -1) ^ ((p-1)/2+ (q-1)/2)that is, two times the reciprocal law was established.

The two reciprocal laws were first presented by Euler and were first proved by Gauss at the age of 19, and now, two reciprocal laws have been proved to be more than ten, which is the pioneering of the century number theory.
So, today I finished two logs in a very noisy environment, tired into DOG... To commemorate the beginning of my little career in gold ...

Finish......

Legendre symbol Correlation