How do i solve this equation using Quadratic reciprocity?
How many solutions does the quadratic equation $\bar{x}^{2} = \bar{2}$ have in $\mathbb{Z}_{47}$?
I have no idea how to go about this i understand if they are in the legrendre symbol format e.g. $$\big(\frac{2}{47}\big)$$ However the squared is confusing me can i put it in to this format?
On
Recall that $2$ is a quadratic residue of the odd prime $p$ if and only if $p\equiv \pm 1\pmod{8}$.
Our prime $47$ is congruent to $-1$ modulo $8$, so $2$ is a quadratic residue of $47$. Thus $x^2\equiv 2\pmod{47}$ has at least one solution.
If $p$ is an odd prime, and $p$ does not divide $a$, then the congruence $x^2\equiv a\pmod{p}$ has no solutions or two solutions. So in our case there are two solutions.
Remark: I would not call the basic facts about $(-1/p)$ or $(2/p)$ quadratic reciprocity.
Using the "Ergänzungssatz" (supplement) of quadratic reciprocity we have $$ \Big( \frac{2}{47}\Big) =(-1)^{\frac{47^2-1}{8}}=(-1)^{276}=1. $$ This says we have a solution. Since $\mathbb{Z}/47$ is a field, the equation $x^2=2$ has exactly two solutions. Indeed, we have $(x^2-2)=(x+7)(x-7)$ in $\mathbb{Z}/47$.