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The relation

$$\left(\frac pq\right)\left(\frac pq\right)=(-1)^{(p-1)/2\cdot(q-1)/2},$$

connecting the Legendre symbols (cf. Legendre symbol)

$$\left(\frac pq\right)\quad\text{and}\quad\left(\frac qp\right)$$

for different odd prime numbers $p$ and $q$. There are two additions to this quadratic reciprocity law, namely:

$$\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}$$

and

$$\left(\frac 2p\right)=(-1)^{(p^2-1)/8}.$$

C.F. Gauss gave the first complete proof of the quadratic reciprocity law, which for this reason is also called the Gauss reciprocity law.

It immediately follows from this law that for a given square-free number $d$, the primes $p$ for which $d$ is a quadratic residue modulo $p$ ly in certain arithmetic progressions with common difference $2|d|$ or $4|d|$. The number of these progressions is $\phi(2|d|)/2$ or $\phi(4|d|)/2$, where $\phi(n)$ is the Euler function. The quadratic reciprocity law makes it possible to establish factorization laws in quadratic extensions $\mathbf Q(\sqrt d)$ of the field of rational numbers, since the factorization into prime factors in $\mathbf Q(\sqrt d)$ of a prime number that does not divide $d$ depends on whether or not $x^2-d$ is reducible modulo $p$.

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