I wish to prove that if $\exists x,y \in Z$ such that $x^2 + 5y^2 = p$ then there cannot exist $a, b \in Z$ such t hat $a^2 + 5b^2 = 2p$. Here p is a prime natural number. I kind of see why this might be true. Here are my ideas so far. Any help in showing why this claim is true will be appreciated.
Assume there exists x,y as stated above. We can assume that they are positive. We see that (x,y) are then the only positive integer solutions to $x^2 + 5y^2 = p$. Assume further that there exists a,b as stated above. Once again, we can assume that they both are positive. Then $a^2 + 5b^2 = 2p$ implies $(\frac{a}{\sqrt{2}})^2 + (\frac{b}{\sqrt{2}})^2 = p$. Now I stuck.
First, note $p$ is not $2$, so it's an odd prime. Thus, $p$ could be congruent to $1$ or $3$ modulo $4$. However, note that $x^2 + 5y^2 \equiv x^2 + y^2 \pmod 4$. Since squares are congruent to only $0$ or $1$ modulo $4$, their sum can be congruent to only $0$, $1$ or $2$ modulo $4$. Thus, you have
$$p \equiv 1 \pmod 4 \implies p \equiv 1,5 \pmod 8 \tag{1}\label{eq1A}$$
As such, since $2(1) \equiv 2(5) \equiv 2 \pmod 8$, you have
$$2p \equiv 2 \pmod 8 \tag{2}\label{eq2A}$$
Squares are $0$, $1$ or $4$ modulo $8$. Thus, $5b^2$ is equivalent to $0$, $5$ or $4$ modulo $8$. The possible combinations are $0$, $1$, $4$, $5$ or $6$ modulo $8$. Thus, there's no combination which sums to $2$ modulo $8$, which is required for $2p$, so there's no $a, b \in Z$ such that $a^2 + 5b^2 = 2p$.