Jacobi symbol:$(\frac{7m^2-1}{18m^2+1})=(\frac{25m^2}{18m^2+1})=1$?

Question

By definition of Jacobi symbol, $(\frac{q}{n})=1$ if there exists $x$ such that $x^2 \equiv q \pmod n$ where $n$ is odd. So, $(\frac{7m^2-1}{18m^2+1})$ is equivalent to-

$x^2 \equiv 7m^2-1 \pmod {18m^2+1} \implies x^2+18m^2+1 \equiv 7m^2-1+18m^2+1 \pmod {18m^2+1}$

[adding $18m^2+1$ in both sides],

$\implies x^2+18m^2+1 \equiv 25m^2 \pmod {18m^2+1}$, clearly, $(\frac{25m^2}{18m^2+1})=1$,

we got, $(\frac{25m^2}{18m^2+1})$ from $(\frac{7m^2-1}{18m^2+1})$,

Does this allow us to write$(\frac{7m^2-1}{18m^2+1})=(\frac{25m^2}{18m^2+1})=1$? Can any one explain a bit more elaborately?

Answer 1

If $a\equiv b\pmod n$ then $\left(\dfrac a n\right)=\left(\dfrac b n\right)$.

Now $7m^2-1\equiv 7m^2-1 +(18m^2+1)\equiv 25m^2\pmod{18m^2+1}.$

Therefore $\left(\dfrac{7m^2-1}{18m^2+1}\right)=\left(\dfrac{25m^2}{18m^2+1}\right).$

Furthermore, $25m^2=(5m)^2$, so $\left(\dfrac{25m^2}{18m^2+1}\right)=1$.