Results 1 to 2 of 2

Thread: 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
    3
    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 $\displaystyle a$ must be an odd integer...

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \displaystyle{\left(\frac{a}{p}\right)\left(\frac{ a}{q}\right)=(-1)^{\frac{a-1}{2}\frac{p-1}{2}}(-1)^{\frac{a-1}{2}\frac{q-1}{2}}\left(\frac{p}{a}\right)\left(\frac{q}{a}\ri ght)=$...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: Sep 3rd 2011, 05:18 AM
  2. Equivalence Relation Proof
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: Mar 25th 2011, 08:32 AM
  3. Legendre symbol proof
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Apr 5th 2010, 10:08 AM
  4. Equivalence Relations Proof
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Nov 10th 2009, 03:33 PM
  5. legendre symbol proof
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: Apr 18th 2009, 10:13 PM

/mathhelpforum @mathhelpforum