= Solution
Write a reduced positive definite <binary quadratic form> as $[a,b,c]=ax^2+bxy+cy^2$, with $|b|\leq a\leq c$ and $b\geq0$ on the reduction boundaries. Since $4ac-b^2=20$, reduction gives $3a^2\leq20$, so $a=1$ or two. The middle coefficient is even. For $a=1$, only $b=0$ is possible, yielding $c=5$. For $a=2$, $b=0$ would give a noninteger $c$, while $b=\pm2$ gives $c=3$; the boundary convention selects $b=2$. Thus the two reduced forms are
$$
\boxed{x^2+5y^2,\qquad 2x^2+2xy+3y^2.}
$$
A positive odd integer $n$ is properly represented by some integer form of discriminant $d$ exactly when $b^2\equiv d\pmod{4n}$ has a solution. For a specified proper equivalence class, the more precise criterion is that $[n,b,(b^2-d)/(4n)]$ belong to that class for some such $b$. Indeed, a primitive representing vector extends to an $SL_2(\mathbb Z)$ basis change, making the leading coefficient $n$; conversely that form represents $n$ at the primitive vector $(1,0)$.
For <primes> $p\nmid10$, the existence criterion is $(-20/p)=1$, because the congruence modulo four is compatible with an even middle coefficient. Quadratic reciprocity gives $(-20/p)=(-1/p)(p/5)$, so the split possibilities are $p\equiv1,3,7,9\pmod{20}$. Any proper representing form reduces to one of the two listed forms. Odd values of $x^2+5y^2$ are one modulo four; odd values of $2x^2+2xy+3y^2$ require odd $y$ and are three modulo four. These distinguish the two classes. Consequently
$$
\boxed{p=x^2+5y^2\text{ for a prime }p\ne2,5\iff p\equiv1\text{ or }9\pmod{20}.}
$$
A <representation> of a <prime> is automatically proper: a common divisor of $x,y$ would have its square divide $p$.
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