I am asked to prove that if $R=\mathbb Z[x,y]/(x^2+y^2-5)$ and $p$ is a prime $\equiv3\pmod4$, then $\text{Spec }(R\otimes_{\mathbb Z}\mathbb F_{p^2})$ is isomorphic to the affine line $\mathbb A_{\mathbb F_{p^2}}^1$. However, I get that it is isomorphic to $\mathbb A_{\mathbb F_{p^2}}^1\setminus\{0\}$. Is my reasoning wrong?
First of all $R\otimes_{\mathbb Z}\mathbb F_{p^2}\simeq\mathbb F_{p^2}[x,y]/(x^2+y^2-5)$. We know that $\mathbb F_{p^2}$ contains a root of $-1$, say $a$. Then we can write $$x^2+y^2-5=(x+ay)(x-ay)-5 =5(\tfrac15(x+ay)(x-ay)-1)$$ and after the change of variables $t=\tfrac15(x+ay)$, $s=x-ay$ we get that $$\mathbb F_{p^2}[x,y]/(x^2+y^2-5)\simeq\mathbb F_{p^2}[s,t]/(st-1)\simeq\mathbb F_{p^2}[s,s^{-1}].$$ Finally, it is well-known that $\text{Spec }\mathbb F_{p^2}[s,s^{-1}]\simeq\mathbb A_{\mathbb F_{p^2}}^1\setminus\{0\}$, so we get $\text{Spec }\mathbb F_{p^2}[x,y]/(x^2+y^2-5)\simeq\mathbb A_{\mathbb F_{p^2}}^1\setminus\{0\}$. (If this works, then it should work also for primes $p\equiv1\pmod4$, $p\neq5$. The only difference is that in that case $a$ is already in $\mathbb F_p$).