Results 1 to 2 of 2

Math Help - Euler/Legendre equivalence proof.

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    1

    Euler/Legendre equivalence proof.

    I would like to show that Euler's and Legendre's formulations of the Law of Quadratic Reciprocity are equivalent:

    Where Euler's version of the law of quadratic reciprocity is
    (a/p) = (a/q),
    with p and odd prime, a not divisible by p, and q a prime with p = +- q (mod 4a).

    And Legendre's version is
    (p/q)(q/p) = (-1) ^ ((p-1)/2)*((q-1)/2),
    where p and q are distinct odd primes.

    How do I approach this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by meetle View Post
    I would like to show that Euler's and Legendre's formulations of the Law of Quadratic Reciprocity are equivalent:

    Where Euler's version of the law of quadratic reciprocity is
    (a/p) = (a/q),
    with p and odd prime, a not divisible by p, and q a prime with p = +- q (mod 4a).

    And Legendre's version is
    (p/q)(q/p) = (-1) ^ ((p-1)/2)*((q-1)/2),
    where p and q are distinct odd primes.

    How do I approach this?
    I think you forgot to mention that a must be an odd integer...

    ---- Suppose Euler's version holds, and let p>q be two odd primes.

    (i) Assume first that p=q\!\!\pmod 4 and put \displaystyle{a:=\frac{p-q}{4} , then:

    \displaystyle{\left(\frac{p}{q}\right)=\left(\frac  {p-q}{q}\right)=\left(\frac{a}{q}\right)\,,\,\,\left(  \frac{q}{p}\right)=\left(\frac{(-1)(p-q)}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{p-q}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)

    Since p=q\!\!\pmod {4a} , we have that \displaystyle{\left(\frac{a}{p}\right)=\left(\frac  {a}{q}\right)\Longrightarrow \left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=\  left(\frac{-1}{p}\right) , and this

    is just Legendre's version for p=q\!\!\pmod 4 (why?) .

    (ii) Now assume p=-q\!\!\pmod 4 and do exactly as above. This time though we get

    that \displaystyle{\left(\frac{a}{p}\right)=\left(\frac  {a}{-q}\right) , and since a=x^2\!\!\pmod q\iff a=x^2\!\!\pmod{-q} , we

    have that \displaystyle{\left(\frac{a}{-q}\right)=\left(\frac{a}{q}\right)\Longrightarrow \left(\frac{q}{p}\right)=\left(\frac{p}{q}\right) , and again this is Legendre's version (why?)

    ---- Suppose now Legendre's version holds , and let p=q\!\!\pmod{4a} , for odd primes p,q, as:

    1) \displaystyle{q=p\!\!\pmod a\Longrightarrow \left(\frac{p}{a}\right)\left(\frac{q}{a}\right)=\  left(\frac{p}{a}\right)\left(\frac{p}{a}\right)=1 ;

    2) Since also p=q\!\!\pmod 4\Longrightarrow\displaystyle{\frac{p-1}{2}+\frac{q-1}{2}=0\!\!\pmod 2 , and thus:

    \displaystyle{\left(\frac{a}{p}\right)\left(\frac{  a}{q}\right)=(-1)^{\frac{a-1}{2}\frac{p-1}{2}}(-1)^{\frac{a-1}{2}\frac{q-1}{2}}\left(\frac{p}{a}\right)\left(\frac{q}{a}\ri  ght)=...now try to complete the

    proof (two steps more). Where did we use that a is an odd integer?

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. equivalence relation proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 3rd 2011, 06:18 AM
  2. Equivalence Relation Proof
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: March 25th 2011, 09:32 AM
  3. Legendre symbol proof
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: April 5th 2010, 11:08 AM
  4. Equivalence Relations Proof
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: November 10th 2009, 04:33 PM
  5. legendre symbol proof
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: April 18th 2009, 11:13 PM

/mathhelpforum @mathhelpforum