This is FKMO 2014 P5.

i) Firstly, there exists $k$ such that $k^{2}\equiv -5\pmod{p}$ ii) According to the Thue's lemma, there exists $x, y$ such that $0 < x, y < \sqrt{p}$ (since $p$ is a prime) satisfy $x \equiv ky\pmod{p}$ This implies $x^{2} \equiv k^{2}y^{2} \equiv -5y^{2}\pmod{p}$ or $p\mid x^{2} + 5y^{2}$ iii) $\frac{x^{2} + 5y^{2}}{p} \in \{1, 2, 3, 4, 5\}$ since $0 < x^{2} + 5y^{2} < 6p$ a) $x^{2} + 5y^{2} = 5p \implies 5\mid x \implies 5x'^{2} + y^{2} = p$. Apply Brahmagupta's identity, the conclusion follows as desired. b) $x^{2} + 5y^{2} = 4p$. Notice that if both $x$ and $y$ are odd then $4p \equiv x^{2} + 5y^{2} \equiv 2\pmod{4}$, which is impossible. Therefore, both $x$ and $y$ are even. Thus, $p = x'^{2} + 5y'^{2}$. c) $x^{2} + 5y^{2} = 3p$ (take note that $\gcd(x, y) = 1$) This implies $(x \pm 5y)^{2} + 5(x \mp y)^{2} = 9p^{2}$. From $\gcd(x, y) = 1$, we get one of $x + 5y$ and $x - 5y$ is divisible by $3$. Thus, the conclusion follows as desired. d) $x^{2} + 5y^{2} = 2p$ implies that $\gcd(x, y) = 1$. Similar to (c), the conclusion follows as desired. e) $x^{2} + 5y^{2} = p$. It's obvious. Thus, we are done. $\bigstar$

so disgustingly routine