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
$$4{x}^{2}+4xy+({2}^{r-2}+1){y}^{2}$$ |
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}$$ |
is reduced and has discriminant $-4n$, so ${h}_{-4n}>1$. So we are left with the case where $n+1$ is a prime power which, since it is even, must be ${2}^{s}$. $s=1,2,3$ correspond to $n=1,3,7$; $s=4$ corresponds to $n=15$, which is not a prime power; and for $s=5$, one can simply compute the forms of discriminant $-4\cdot 31$ to see that ${h}_{-4\cdot 31}=3$. So the only possibility remaining is that $s\ge 6$. In this case, though,
$$8{x}^{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. ∎
References
- 1 Cox, D.A. Primes of the Form ${x}^{\mathrm{2}}\mathrm{+}n\mathit{}{y}^{\mathrm{2}}$: Fermat, Class Field Theory, and Complex Multiplication^{}, Wiley 1997.
Title | there is a unique reduced form of discriminant $-4n$ only for $n=1,2,3,4,7$ |
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Canonical name | ThereIsAUniqueReducedFormOfDiscriminant4nOnlyForN12347 |
Date of creation | 2013-03-22 16:56:46 |
Last modified on | 2013-03-22 16:56:46 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 6 |
Author | rm50 (10146) |
Entry type | Theorem |
Classification | msc 11E12 |
Classification | msc 11R29 |
Classification | msc 11E16 |