Qfyilyi

Let $ax^2 + bx + c = 0$ and $dy^2 + ey + f = 0$ such that variables $a$ through $f$ are fixed integer parameters. I'm trying to find integers $g$, $h$, and $j$ such that $g(xy)^2 + h(xy) + j = 0$. I imagine the equation will be different depending on which roots of the original polynomials we consider. I'm looking for a proof either they exist or don't exist and if they do exist what they are in terms of the original integer values.

The motivation is that I'm trying to represent certain numbers using the form $x = [S, A, B, C]$ where $S$ denotes which root $x$ is to a quadratic polynomial $Ax^2 + Bx + C$. I'm trying to figure out how to multiply such numbers.

Here's one way to understand why such $g,h,j$ don't necessarily exist:

The roots of an integer quadratic are expressible as

$$frac{xpm sqrt{y}}{z}$$

for some integers $x,y,z$ (say, by the quadratic formula). If you multiply two of these together, you can get something like this:

$$left(frac{x_1+sqrt{y_1}}{z_1}right)left(frac{x_2+sqrt{y_2}}{z_2}right)=frac{x_1x_2+x_1sqrt{y_2}+y_1sqrt{x_2}+sqrt{y_1y_2}}{z_1z_2}.$$

We need these three square-root terms to somehow collapse into one for this to be expressible as $frac{X+sqrt{Y}}{Z}$ for some integers $X,Y,Z$. If there are no nontrivial relations between $sqrt{y_1}$ and $sqrt{y_2}$ (say, if $y_1=2$ and $y_2=3$), then we can't simplify the above at all, and thus it isn't a root of an integer-coefficient quadratic.

There is a generic polynomial that will get you the products of the roots - but you can't keep the degree the same. After all, there's not necessarily any way to distinguish between the roots, so if one product appears as a root, all possible products of roots do. For two quadratics, that means that we're looking at a fourth-degree polynomial.

For simplicity, I'll work with two monic quadratics, $x^2+ax+b$ and $x^2+cx+d$, with roots $r_1,r_2$ and $s_1,s_2$ respectively. We have $a=-(r_1+r_2)$, $b=r_1r_2$, $c=(-s_1+s_2)$, $d=s_1s_2$, and we want to find the symmetric functions of the products of roots to get the quartic $x^4+ex^3+fx^2+gx+h$.

First, the constant term $h=(r_1s_1)(r_1s_2)(r_2s_1)(r_2s_2)=r_1^2r_2^2s_1^2s_2^2=b^2d^2$.

Second, the $x^3$ term $e=-(r_1s_1+r_1s_2+r_2s_1+r_2s_2)=-(r_1+r_2)(s_1+s_2)=-ac$.

Third, the $x$ term
begin{align*}g &=-r_1s_1r_1s_2r_2s_1-r_1s_1r_1s_2r_2s_2-r_1s_1r_2s_1r_2s_2-r_1s_2r_2s_1r_2s_2\
&=-r_1r_2s_1s_2(r_1s_1+r_1s_2+r_2s_1+r_2s_2)\
&= -abcdend{align*}
Finally, the $x^2$ term
begin{align*}f &= r_1s_1r_1s_2+r_1s_1r_2s_1+r_1s_1r_2s_2+r_1s_2r_2s_1+r_1s_2r_2s_2+r_2s_1r_2s_2\
&= (r_1^2+r_2^2)s_1s_2+r_1r_2(s_1^2+s_2^2)+2r_1r_2s_1s_2\
&= (a^2-2b)d+(c^2-2d)b+2bd = a^2d+c^2b-2bdend{align*}
Putting those together, the final polynomial is
$$x^4 -ac x^3 + (a^2d+c^2b-2bd)x^2 -abcd x + b^2d^2$$

If we tried to do the same with fewer of the products, to build a quadratic, it just wouldn't work. The building blocks we have are the symmetric polynomials in $r_1,r_2$ and $s_1,s_2$, so anything we build with them will have to be symmetric as well; if $r_1s_1$ is a root, we can exchange $r_1$ and $r_2$ to get $r_2s_1$ as another root, or exchange $s_1$ and $s_2$ to get $r_1s_2$ as a third root, or exchange both to get $r_2s_2$ as a fourth root. It's all four or none. We've shown that getting just those four roots is possible, so that's the best that can be done.

In fact, it's typical that for two irreducible quadratics, the quartic we build this way will be irreducible as well. We'd need a coincidence like the discriminants being the same (up to multiplication by a perfect square) to avoid it.