Do we have a fast method for proving that the equation $x^2+y^2=0$ over $\mathbb{F}_{7}$ has one solution $(0,0)$ without testing all the elements of $\mathbb{F}_{7}$?
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Try testing $x^2\operatorname{mod}7$ from $x=1$ to $3$. – Kemono Chen Dec 21 '19 at 13:21
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2Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Dec 21 '19 at 13:26
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This could help ? https://en.wikipedia.org/wiki/Quadratic_reciprocity#q_=_%C2%B11_and_the_first_supplement – Donald Splutterwit Dec 21 '19 at 13:37
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Yes, at least for prime fields – using quadratic reciprocity:
Indeed, if the equation $x^2+y^2=0$ has a solution such that $x,y\ne 0$, then $-1$ is a square in $\mathbf F_p$. Now, the $1$st supplementary law of quadratic reciprocity asserts that $$\biggl(\frac{-1}{p}\biggr)=\bigl(-1\bigr)^{\tfrac{p-1}2}, $$ and if $p=7$, this is equal to $-1$.
Bernard
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To answer the general question suggested in the title, consider a finite field $\mathbb F_q$. Then:
$x^2+y^2=0$ has a nontrivial solution in $\mathbb F_q$ iff $z^2+1=0$ has a solution in $\mathbb F_q$ iff there is an element of order $4$ in $\mathbb F_q^\times$ iff $4$ divides $q-1$, since $\mathbb F_q^\times$ is cyclic.
lhf
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