Let $p$ be a prime number. We claim that $x^2 + y^2 = p$ has a rational solution if and only if $p = 2$ or $p \equiv 1$ (mod 4).We assume we have the basic knowledge of the ideal theory of $\mathbb{Z}[\sqrt{-1}]$.
Let $K = \mathbb{Q}(\sqrt{-1})$. Suppose $x^2 + y^2 = p$ has a rational solution. Let $\gamma = x + \sqrt{-1}y \in K$. Then $N(\gamma) = p$. Let $A = \mathbb{Z}[\sqrt{-1}]$. It is well known that $A$ is a principal ideal domain. Suppose $\gamma = \alpha/\beta$, where $\alpha, \beta \in A$ and $\alpha$ and $\beta$ are relatively prime. Then $N(\gamma) = N(\alpha)/N(\beta)$. Hence $N(\alpha) = pN(\beta)$. Since $\alpha$ and $\beta$ are relatively prime, $N(\alpha)$ and $N(\beta)$ are relatively prime. Hence $N(\beta)$ is not divisible by $p$. Since $N(\alpha)$ is divisible by $p$, there exists a prime ideal $P$ dividing $\alpha$ and $p$. Then $N(P) = p$. Hence $P \neq pA$. Hence, by the theorem of the decomposition of a prime number in $K$, $p = 2$ or $p \equiv 1$ (mod 4).
Conversely suppose $p = 2$ or $p \equiv 1$ (mod 4). There exists a prime ideal $P$ such that $N(P) = p$. Hence $x^2 + y^2 = p$ has an integer solution. QED