Why is the Kronecker symbol $(n/2)$ conventionally defined as it is?

Question

Admittedly this may be an extremely naive question, but I am simply puzzled about the motivation behind choosing this function to be one of the Dirichlet characters modulo $8$, rather than the Dirichlet character modulo $2$ (or, indeed, any other character modulo $8$). There must be a good reason for this, right? What was Kronecker's motivation?

For $m\in\mathbb{N}$ and $n\notin 2\mathbb{N}$, the Kronecker symbol $(m/n)$ is the completely multiplicative extension to the odd numbers of the Legendre symbol $(m/p)$, where $p$ is an odd prime. That is,

$$\left(\frac{m}{n}\right)=\prod_{p|n}\left(\frac{m}{p}\right)^{\alpha_p(n)},$$ where $\alpha_p(n)$ is the exponent of $p$ in the factorisation of $n$ and $\left(\frac{m}{p}\right)$ is $0$ if $p|m$ and $\{-1,1\}$ depending on whether $m$ is a quadratic non-reside or residue (mod $p$), respectively.

Very well. When $n\in 2\mathbb{N}$, however, Kronecker's symbol is defined in the same way but with $(m/2)$ defined to be $0$ if $m$ is even, $1$ if $m\equiv \pm 1$ (mod 8) and $-1$ $m\equiv \pm 3$ (mod 8). In other words, Kronecker's symbol the Dirichlet character modulo $8$ that takes the values $1,0,-1,0,-1,0,1,0,...$.

Of course, there are four characters (mod $8$) which are all real, completely multiplicative functions of $m$. For example, the simplest of these is the Dirichlet character (mod $2$), which takes the values $1,0,...$. In particular, had we taken Kronecker's symbol to be this character, then it would still be completely multiplicative as a function of $n$ and when $n=2$ we would have a function of $m$ that carries the trivial structure of residuosity (mod $2$), too. So why do we take it to be the particular character we do, instead?

Answer 1

The reason is quadratic residues and Legendre's symbol $\,\big(\frac{a}{p}\big).\,$ In particular the Legendre symbol $\,\big(\frac2{p}\big).\,$ It can be proved that $2$ is a quadratic residue of primes of the form $\,8n+1\,$ or $\,8n+7\,$ and a quadratic nonresidue of primes of the form $\,8n+3\,$ or $\,8n+5.\,$ This can be expressed by the formula $\,\big(\frac2{p}\big) = (-1)^{(p^2-1)/8}\,$ which is contained in the Wikipedia article on Legendre symbol.

Answer 2

The "correct" definition of the Kronecker symbol is the following. Let $D$ be the discriminant of a quadratic number field and $p$ a nonramified prime. Then $(D/p) = +1$ or $-1$ according as $p$ splits or is inert. Thus the Kronecker symbol encodes the splitting of primes in quadratic extensions (which is what the Artin symbol does in arbitrary abelian extensions). The extension to composite values of $p$ is achieved by multiplicativity.

If $K = {\mathbb Q}(\sqrt{2})$, then $D = 8$, and $(D/p) = +1$ if $p \equiv 1 \bmod 8$, and $(D/p) = -1$ if $p \equiv 5 \bmod 8$.