there is a unique reduced form of discriminant $-4n$ only for $n=1,2,3,4,7$

The number of reduced http://planetmath.org/node/IntegralBinaryQuadraticFormsintegral binary quadratic forms of a given http://planetmath.org/node/IntegralBinaryQuadraticFormsdiscriminant is finite; the number of such forms is $h_{\mathrm{\Delta }}$.

Theorem 1.

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

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 $x^2+n{y}^2$ 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 . Then $a{x}^2+c{y}^2$ 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 $r\ge 4$, then

$$4x^2+4xy+(2^{r-2}+1)y^2$$

is clearly primitive, and is also reduced since $4\le 2^{r-2}+1$. 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 . Then

$$a{x}^2+2xy+c{y}^2$$

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 $s\ge 6$. In this case, though,

$$8x^2+6xy+(2^{s-3}+1)y^2$$

has relatively prime coefficients, and is reduced since $s\ge 6\Rightarrow 8\le 2^{s-3}+1$. Also, its discriminant is $6^2-4\cdot 8\cdot (2^{s-3}+1)=4-4\cdot 2^s=4-4(n+1)=-4n$, and thus $h_{-4n}>1$ in this case as well.∎