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


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.
