Theorem 1   Let $n$ be a positive integer. Then $h_{-4n}=1$ if and only if $n=1,2,3,4,{ or }7$ .

Proof. (This proof is taken from [1], which is itself taken from an earlier proof).
By computing all reduced forms of discriminants $-4,-8,-12,-16, -20$ one can see that $h_{-4n}=1$ in the five cases given in the statement of the theorem. We show there are no others.

Clearly $x^2+ny^2$ is a reduced form of discriminant $-4n$ . For $n\notin \{1,2,3,4,7\}$ , we will produce a second reduced form of the same discriminant, showing that $h_{-4n}>1$ . We may assume $n>1$ , since we already know that $h_{-4}=1$ .

Suppose first that $n$ has at least two distinct prime factors. Then we can write $n=ac$ where $1<a<c,\ \gcd(a,c)=1$ . Then $ax^2+cy^2$ is reduced, and its discriminant is $-4ac=-4n$ . So if $n$ has two distinct prime factors, $h_{-4n}>1$ .

We now consider the prime power case, taking $2^r$ and $p^r$ , $p$ an odd prime, separately.

If $n=2^r$ , then we already know that for $r=1,2$ , $h_{-4n}=1$ . For $r=3$ , one can compute the classes of discriminant $-32$ and see that $h_{-4n}=2$ . For $r\geq 4$ , then $$ 4x^2+4xy+(2^{r-2}+1)y^2 $$ is clearly primitive, and is also reduced since $4\leq 2^{r-2}+1$ . Further, its discriminant is $4^2-4\cdot 4\cdot(2^{r-2}+1)=-16\cdot 2^{r-2}=-4n$ . Thus in this case as well, $h_{-4n}>1$ .

Finally, suppose $n=p^r$ , $p$ an odd prime. Suppose we can write $n+1=ac, 2\leq a<c,\ \gcd(a,c)=1$ . Then $$ ax^2+2xy+cy^ $$ is reduced and has discriminant $-4n$ , so $h_{-4n}>1$ . So we are left with the case where $n+1$ is a prime power which, since it is even, must be $2^s$ . $s=1,2,3$ correspond to $n=1,3,7$ ; $s=4$ corresponds to $n=15$ , which is not a prime power; and for $s=5$ , one can simply compute the forms of discriminant $-4\cdot 31$ to see that $h_{-4\cdot 31}=3$ . So the only possibility remaining is that $s\geq 6$ . In this case, though, $$ 8x^2+6xy+(2^{s-3}+1)y^ $$ has relatively prime coefficients, and is reduced since $s\geq 6\Rightarrow 8\leq 2^{s-3}+1$ . Also, its discriminant is $6^2-4\cdot 8\cdot(2^{s-3}+1)=4-4\cdot2^s=4-4(n+1)=-4n$ , and thus $h_{-4n}>1$ in this case as well.