= Solution
An integral positive definite <binary quadratic form>
$$
[a,b,c](x,y)=ax^2+bxy+cy^2
$$
is <reduced positive definite binary quadratic form>[reduced] when
$$
|b|\le a\le c,
$$
with $b\ge0$ when $|b|=a$ or $a=c$. Its <discriminant of a binary quadratic form>[discriminant] is $D=b^2-4ac$.
For $D=-20$, reduction gives $3a^2\le20$, so $a=1$ or $2$. Since $b^2\equiv D\pmod4$, $b$ is even. If $a=1$, then $|b|\le1$ forces $b=0$, and $c=5$. If $a=2$, then $b\in\{-2,0,2\}$; integrality of
$$
c=\frac{b^2+20}{8}
$$
leaves $b=\pm2$ and $c=3$, and the boundary convention retains only $b=2$. Hence the complete list is
$$
\boxed{[1,0,5]\quad\text{and}\quad[2,2,3]}.
$$
Now suppose the prime $p\ne5$ is represented by $x^2+5y^2$. If $p=2$, the equation is impossible modulo four, so $p$ is odd. The integers $x$ and $y$ then have opposite parity, and therefore
$$
p=x^2+5y^2\equiv1\pmod4.
$$
Also $x\not\equiv0\pmod5$, since otherwise $5\mid p$; hence
$$
p\equiv x^2\equiv1\ \text{or}\ 4\pmod5.
$$
The <Chinese remainder theorem> gives the stronger conclusion
$$
p\equiv1\ \text{or}\ 9\pmod{20}.
$$
In particular,
$$
\boxed{p\equiv1,3,7,\text{ or }9\pmod{20}},
$$
as required.
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