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Only for illustration, let's choose an integer $N=6$. I want to create a collection of all possible quadratic equations in the form of $ax^2+bx+c=0$ where

  • $|ac|=|\alpha\beta|=N=6$, and
  • $\alpha+\beta=b$.

Note that all constants are integers but $x$ is not necessarily integer.

Here is my attempt.

How many ways to specify $a$ and $c$

The prime factors of $N$ is $$ P=\{2,3\} $$

The all possible pairs of $(a,c)$ are $$ \{(1,\pm 6), (6,\pm 1), (2,\pm 3), (3,\pm 2)\} $$ Here I don't need to consider the case in which $a<0$ because, for example, $6x^2+5x-1=0$ is actually identical to $-6x^2-5x+1=0$. Only the sign of $b$ and $c$ do matter here. From $2^2=4$ subsets of $P$, we can create $\frac{2^2}{2}=2$ "partitions", each "partition" can be assigned to $(a,c)$ in $2!=2$ ways, and $c$ can have two choice of signs. Thus there are $$ \frac{2^2}{2}\times 2! \times 2 = 8 $$ ways to create $(a,c)$.

How many ways to specify $b$

The all possible pairs of $(\alpha,\beta)$ are $$ \{(\pm 1,\pm 6), (\pm 2,\pm 3)\} $$

Therefore the all possible values of $b$ are $$ \{\pm |1+6|, \pm |1-6| = \pm |2+3|, \pm |2-3|\} = \{\pm 7, \pm 5, \pm 1 \} $$

I cannot find an easier way to calculate how many possible ways to determine $b$ because the same $b$ can be obtained from two (or possibly more) "partitions". For example, partition $\{1,6\}$ and $\{2,3\}$ can produce $b=\pm 5$ as follows. $$ \pm |1-6| = \pm |2+3| $$

As we can see, there are $3\times 2=6$ ways to assign $b$.

Final

Thus in total, we have $8\times 6=48$ quadratic equations. I have not checked programmatically whether all of these equations have distinct pair of roots.

Question

Generally speaking, for any positive integer $N$, how many quadratic equations (with the constraints given above) are possible to make?

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  • 1
    $\begingroup$ I guess $a$, $b$, $\alpha$, and $\beta$ should be integers. $\endgroup$ – Alex Ravsky Sep 18 '20 at 0:09
  • $\begingroup$ It seems to me that you are not saying that $a,b,c,\alpha,\beta$ are integers. If $a,b,c,\alpha,\beta$ are not necessarily integers, then there are infinitely many such quadratic equations. Take $(a,b,c,\alpha,\beta)=\left(\sqrt{\frac{4N}{2m-1}},ma,\frac Na,\frac{b-\sqrt{b^2+4N}}{2},\frac{b+\sqrt{b^2+4N}}{2}\right)$ where $m$ is a rational number larger than $1$. This satisfies $|ac|=|\alpha\beta|=N$ and $\alpha+\beta=b$, and the roots of $ax^2+bx+c=0$ are $x=-\frac 12,\frac{1-2m}{2}$. $\endgroup$ – mathlove Sep 19 '20 at 5:53
  • $\begingroup$ @AlexRavsky: Yes. They are integers as shown in my attempt. $\endgroup$ – Artificial Stupidity Sep 19 '20 at 6:53
  • $\begingroup$ @mathlove: They are integers, except for $x$ that is not necessarily integer. $\endgroup$ – Artificial Stupidity Sep 19 '20 at 6:55
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This is a partial answer.

This answer deals with $N$ such that $N\not\equiv 0\pmod 3$.

This answer proves that the number of such quadratic equations is $$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$} \\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$ where $\sigma_0(N)$ is the number of the positive divisors of $N$.

Proof :

We may suppose that $a\gt 0$.

Note that $ax^2+bx+c=0$ has two distinct rational roots if and only if $b^2-4ac$ is a non-zero square number.

Also, let $\sigma_0(N)$ be the number of the positive divisors of $N$.

($\sigma_0(N)=\displaystyle\prod_{k=1}^{d}(e_k+1)$ when $N=\displaystyle\prod_{k=1}^{d}p_k^{e_k}$ where $p_1,p_2,\cdots,p_k$ are distinct prime numbers.)

  • Case 1 : $ac=\alpha\beta$

    Since we have $b^2-4ac=(\alpha+\beta)^2-4\alpha\beta=(\alpha-\beta)^2$, we see that $ax^2+bx+c=0$ has two distinct rational roots if and only if $\alpha\not=\beta$.

    If $\alpha\beta=\alpha'\beta',\alpha'\not=\alpha$ and $\alpha'\not=\beta$, then $$\small\alpha+\beta-(\alpha'+\beta')=\alpha+\beta-\alpha'-\frac{\alpha\beta}{\alpha'}=\frac{(\alpha-\alpha')(\alpha'-\beta)}{\alpha'}\not=0\implies \alpha+\beta\not=\alpha'+\beta'\tag1$$Case 1-1 : $ac=\alpha\beta=N$

    The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.

    If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.

    If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2(\sigma_0(N)-1)$ since the case where $\alpha=\beta$ has to be excluded. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)-1$. So, the number of such quadratic equations is $\sigma_0(N)(\sigma_0(N)-1)$.

    Case 1-2 : $ac=\alpha\beta=-N$

    The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.

    If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.

    If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$ since the case where $\alpha=\beta$ does not happen. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.

  • Case 2 : $ac=-\alpha\beta=\pm N$

    For $N$ such that $N\not\equiv 0\pmod 3$, we have $$\begin{align}-\alpha\beta=\pm N&\implies (\alpha,\beta)\equiv (1,1),(1,2),(2,1),(2,2)\pmod 3 \\\\&\implies b^2-4ac=(\alpha+\beta)^2+4\alpha\beta\equiv 2\pmod 3\end{align}$$So, $b^2-4ac$ cannot be a square number.

Hence, it follows from the above cases that the number of such quadratic equations is $$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$} \\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$ where $\sigma_0(N)$ is the number of the positive divisors of $N$.

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