For $a\in\mathbb{Z}$ that is not a perfect square such that $a\equiv0\text{ or }1\left(\mod4\right)$, define $$\left(\frac{a}{2}\right):=\begin{cases} 1 & \text{if }a\equiv1\left(\mod8\right)\\ -1 & \text{if }a\equiv5\left(\mod8\right)\\ 0 & \text{if }a\equiv0\left(\mod4\right) \end{cases}$$ Moreover, for any $n\in\mathbb{N}\backslash\left\{ 1\right\}$ with $\left(a,n\right)=1$, define: $$\left(\frac{a}{n}\right):=\left(\frac{a}{2}\right)^{e}\prod_{j=1}^{r}\left(\frac{a}{p_{j}}\right)^{e_{j}},\text{if }n=2^{e}\prod_{j=1}^{r}p_{j}^{e_{j}}\text{ (${p_{j}:}$odd) is the prime factorization of }n.$$Let a be as above. Prove that if $n_{1},n_{2}\in\mathbb{N}\backslash\left\{ 1\right\}$ such that $\left(a,n_{1}\right)=\left(a,n_{2}\right)=1$ and $n_{1}\equiv n_{2}\left(\mod\left|a\right|\right)$, then $$\left(\frac{a}{n_{1}}\right)=\left(\frac{a}{n_{2}}\right).$$
Thoughts: Since $n_1$ and $n_2$ can be even, I cannot use Jacobi symbol directly. I have to use what is defined above. I have seen similar problems for this, however the "numerator" (I don't know how to call it) is the one with conguency not the "denominator". So I f have to use a similar argument, I need to figure out how to separate $n_{1}\equiv n_{2}\left(\mod\left|a\right|\right)$ into $p_{j}\equiv n_{i}\left(\mod\left|a_j\right|\right)$. I don't if I'm doing it right. Any help would be much appreciated. Thank you.