How to show that either $a^2+5b^2=p$ or $c^2+5d^2=2p$ has integer solutions for all prime $p$ with $(\frac{-5}{p})=1$ How to show that either $a^2+5b^2=p$ or $c^2+5d^2=2p$ has integer solutions for all prime $p$ with $(\frac{-5}{p})=1$
would the fact that $\mathbb{Z}[\sqrt{-5}]$=$\mathcal{O}\cap \mathbb{Q}[\sqrt{-5}]$ has class number $2$ be any use.
 A: Looking at the possible values of $a^2+5b^2 \pmod {5}$, we find that for any prime $p \neq 5$, 
$a^2 + b^2 = p \implies p \equiv 1,4 \pmod 5$ and $a^2 + b^2 = 2p \implies p \equiv 2,3 \pmod 5$
If $-5$ is a square modulo $p$, then $(p)$ splits in $\Bbb Z[\sqrt{-5}]$ into a product $\mathfrak p\bar{\mathfrak p}$, where $\mathfrak p$ has norm $p$.
Since $\Bbb Z[\sqrt{-5}]$ has class number $2$, we have that for any such prime,
$$Cl(\mathfrak p) = Cl(\Bbb Z[\sqrt{-5}]) \iff \exists y \in \Bbb Z[\sqrt{-5}], \mathfrak p = y\Bbb Z[\sqrt{-5}] \implies \\
\exists y \in \Bbb Z[\sqrt{-5}], p = N(y) \iff \exists a,b \in \Bbb Z, a^2+5b^2 = p \implies p \equiv 1,4 \pmod 5$$
and since the ideal $I = \langle 2, 1+\sqrt{-5}\rangle$ is a nonprincipal ideal of norm $2$, $I^2$ is principal, and :
$$Cl(\mathfrak p) = Cl(I) \iff Cl(I)Cl(\mathfrak p) = Cl(\Bbb Z[\sqrt{-5}]) \iff \exists y \in \Bbb Z[\sqrt{-5}], \mathfrak pI = y\Bbb Z[\sqrt{-5}] \implies \\
\exists y \in \Bbb Z[\sqrt{-5}], 2p = N(y) \iff \exists a,b \in \Bbb Z, a^2+5b^2 = 2p \implies p \equiv 2,3 \pmod 5$$
Once we realise that both the starting points and the ending points are mutually exclusive, and that the starting points cover all possible cases, we can replace the implications in the middle with equivalences.

Also, the only way for $a^2+5b^2$ to be even is to have $a$ and $b$ of the same parity.  If you replace $a$ with $2c+b$, you obtain that for any integer $p$, $\exists a,b \in \Bbb Z, a^2+5b^2 = 2p \iff \exists b,c \in \Bbb Z, 2c^2+2bc+3b^2 = p$. This formulation is cleaner : it doesn't use the prime $2$ anymore (why use $2p$ rather than $3p$ or $7p$ or any other ?) and we are left with the two non-equivalent quadratic forms of discriminant $-20$
A: The fact that you state doesn't seem to be of any relevance to me.
A hint to a solution: What can you say about $-5$ when $\left(\frac{-5}p\right)=1$? And how can you use that to find integer solutions for $a^2+5b^2=p$?
