The denominator of the rational roots

Introduction

In the previous post, I factorised a quadratic equation with a fraction coefficient $x^2+\tfrac16 x-2$ by thinking of numbers that add to 1/6 and multiply to -2. This is how I did it:

2/6+(-1/6)=1/6, 2/6×(-1/6)=1/3×(-1/6)=-1/18, which is not low enough.
3/6+(-2/6)=1/6, 3/6×(-2/6)=1/2×(-1/3)=-1/6, which is lower but not low enough, and I’ve got quite a long way to go.
7/6+(-6/6)=1/6, 7/6×(-6/6)=7/6×(-1)=-7/6, so much closer.
8/6+(-7/6)=1/6, 8/6×(-7/7)=… yeah that won’t work out right.
9/6+(-8/6)=1/6, 9/6×(-8/6)=3/2×(-4/3)=-2 yay!

But how could I be sure that counting in sixths would eventually get me to the right place? What if it was some other denominaror I needed?

Well, it turns out that if you want two numbers with specific rational sum and product, they are guaranteed to be able to be written with a specific denominator, so you will be able to try only numbers with that denominator.

The theorem

Theorem:
Let \(a\), \(b\), \(c\) and \(d\) be integers with \(b\) and \(d\) not zero. Suppose there are rational numbers \(x\) and \(y\) such that \(x+y=\frac{a}{b}\) and \(xy=\frac{c}{d}\). Then it is possible to write both \(x\) and \(y\) with denominator \(bd\).

I would like to show my proof for this theorem. There might be an easier way than I did it, but I definitely enjoyed my way, so I want to share it. It’s actually two proofs, but the first one makes me feel a little uncomfortable. I’ll do that one first.

Proof 1

Since \(x\) and \(y\) are rational, let \(x=\frac{p}{q}\) and \(y=\frac{r}{s}\) for integers \(p\), \(q\), \(r\), and \(s\) with \(q\) and \(s\) not zero.

Then

\[\begin{aligned} x+y&=\frac{p}{q}+\frac{r}{s}\\ &=\frac{ps}{qs}+\frac{qr}{qs}\\ &=\frac{ps+qr}{qs}\end{aligned}\]

and

\[\begin{aligned}xy&=\frac{p}{q}\cdot\frac{r}{s} \\ &= \frac{pr}{qs}\end{aligned}\]

Thus both the sum and the product can be written with denominator \(qs\). The numbers \(x\) and \(y\) themselves can also be written with that denominator, as

\[\begin{aligned} x&= \frac{p}{q} = \frac{ps}{qs}\\ y&= \frac{r}{s} = \frac{qr}{qs}\end{aligned}\]

In other words, the numbers \(x\) and \(y\) can be written with the same denominator as the common denominator of the sum and the product. The sum \(\frac{a}{b}\) and the product \(\frac{c}{d}\) have \(bd\) as a common denominator, so that means \(x\) and \(y\) can be written with this denominator.

Interlude

I am certain this proof is watertight, except for maybe tidying up the idea that if a two fractions can be written with a common denominator, then they can be written with any of the common denominators.

But still it feels a bit uncomfortable somehow. It gives a whiff of circular reasoning, maybe, or at least I don’t directly talk about the original denominators \(b\) and \(d\) anywhere until I reveal them at the last moment, which feels sneaky. I’d much prefer a proof that begins with the equations and ends with solutions with the correct denominators. And so I have this proof too.

Proof 2

Consider the two equations:

\[\begin{aligned}x+y&=\frac{a}{b}\\xy&=\frac{c}{d}\end{aligned}\]

Multiply the first by \(bdx\) and the second by \(d\) to give

\[\begin{aligned}bdx^2+bdxy&=adx\\dxy&=c\end{aligned}\]

Subsitute the second equation into the first to give

\[\begin{aligned}bdx^2+bc=adx\\bdx^2-adx+bc=0\end{aligned}\]

Using completing the square or the quadratic formula,

\[x = \frac{ad\pm\sqrt{(ad)^2-4b^2cd}}{2bd}\]

Now \((ad)^2-4b^2cd\) is an integer, so \(\sqrt{(ad)^2-4b^2cd}\) is either unreal, irrational or an integer. Since \(x\) is rational, that means it must be an integer. Let it be \(m\), so that

\[x = \frac{ad\pm m}{2bd}\]

Suppose \(ad\) is odd. Then \((ad)^2\) is odd, and so is \((ad)^2-4b^2cd\), and therefore so is \(\sqrt{(ad)^2-4b^2cd}=m\). But now \(ad\pm m\) is even.

Alternatively suppose \(ad\) is even. Then \((ad)^2\) is a multple of 4, and so is \((ad)^2-4b^2cd\), which means \(\sqrt{(ad)^2-4b^2cd}=m\) is even. But now again \(ad\pm m\) is even.

Hence \(x\) can be written with an integer numerator and denominator as

\[x = \frac{ad\pm m}{2bd} = \frac{\left(\frac{ad\pm m}{2}\right)}{bd}\]

Because of the symmetry in the original equations, this is also the two solutions for \(y\). Thus both \(x\) and \(y\) can be written with denominator \(bd\).

Conclusion

I particularly enjoyed that second proof. I was convinced when I did it the first time that the denominator had to be \(2bd\) but then I realised that the numerator had to be even and so it came down to \(bd\) after all, which was very satisfying.

And I am very happy that this means I can now factorise monic quadratics with rational coefficients by directly working through numbers with one specific denominator that add to the x-coefficient. It just feels like such a bold move to me, and now I know it just seems bold, because it provably will definitely work.