Why in $GF(7)$ there are no nonzero elements satisfying $a^2 + b^2 = 0$?

Question

Why in $GF(7)$ there are no nonzero elements satisfying $a^2 + b^2 = 0$? how can I know this by a quick method without calculations? Is there a theorem or something that can help here?

$GF(7)$ is the integers modulo $7.$

Answer 1

If $a^2 + b^2 = 0$, and $b \ne 0$, then $(a/b)^2 = -1$, so $a/b$ has multiplicative order $4$. Now the order of any group element divides the order of the group. In this case the group has order $6$. Since $4$ does not divide $6$, we're done.

Answer 2

If there exists $a,b$ non-zero such that $a^{2}+b^{2}\equiv 0\pmod{7}$, then this would mean $$a^{2}\equiv -b^{2}\pmod{7}\implies \left(\frac{-1}{7}\right)=1$$ which is absurd as by Euler's Criterion we have $\left(\frac{-1}{p}\right)=-1$ for primes congruent to $3\pmod{4}$.