# 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 $\Delta<0$ is finite; the number of such forms is $h_{\Delta}$.

###### Theorem 1.

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

###### 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. 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. Then

 $ax^{2}+2xy+cy^{2}$

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

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

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

## References

Title there is a unique reduced form of discriminant $-4n$ only for $n=1,2,3,4,7$ ThereIsAUniqueReducedFormOfDiscriminant4nOnlyForN12347 2013-03-22 16:56:46 2013-03-22 16:56:46 rm50 (10146) rm50 (10146) 6 rm50 (10146) Theorem msc 11E12 msc 11R29 msc 11E16