Let $a\not\equiv 3\pmod 4$ and $a\ne 0$. How do you show that $\chi(n):=\left(\frac{a}{n}\right)$ (where $\left(\frac{\cdot}{\cdot}\right)$ denotes the Kronecker symbol) defines a character of modulus $m:=4|a|$ if $a\equiv 2 \pmod 4$ and $m:=|a|$ otherwise?
What I have to show is that we have $\left(\frac{a}{n}\right)=\left(\frac{a}{m+n}\right)$. The Kronecker symbol is defined using the prime factorisation of the lower argument. Unfortunately the prime factorisation of $n$ and $n+m$ can be totally different. So I don't have a clue how to prove that identity.
The Kronecker symbol defines a character, because it is completely multiplicative by definition, i.e., because the Jacobi symbol is multiplicative. So we have $\chi(nm)=\left(\frac{a}{nm}\right)=\left(\frac{a}{n}\right)\left(\frac{a}{m}\right)=\chi(n)\chi(m)$, unless $a=-1$, one of $m,n$ is zero and the other one has odd part congruent to $3\bmod4$. So the only problem is, to be a bit careful with the different cases of the definition - see here.
It also follows from the properties of the Jacobi symbol that for $a\not\equiv3\pmod 4$, $a\ne 0$, we have $\left(\tfrac am\right)=\left(\tfrac an\right)$ whenever $$m\equiv n\bmod\begin{cases}4|a|,&a\equiv2\pmod 4,\\|a|&\text{otherwise.}\end{cases}$$