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?

2. Originally Posted by meetle
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