Strange Mathematics

Eric Bailey

Written on 2 August, 2020
Tags: , ,

On the first page of Tom King's Strange Adventures #2 there is an example of bad mathematics.

T-Sphere:

IN THE EQUATION
X² + MX + N = 0,
M AND N ARE INTEGERS.
THE ONLY POSSIBLE VALUE
OF X IS -3. WHAT IS THE
VALUE OF M?

Mister Terrific:

SIX.

T-Sphere:

CORRECT.

Update (3 August, 2020): It turns out I'm the one bad mathematics (and sleep hygiene).

Given \(x^2 + mx + n = 0\) and the only possible value of \(x\) is \(-3\), \((x + 3)\) is a root of multiplicity \(2\). Thus the factorization is \((x + 3)(x + 3)\), which expands to \(x^2 + 6x + 9\). Thus \(m = 6\), just as Mister Terrific said. 🤦

But that's not quite right, or at least it's misleading. While \(\xcancel{6}\) is a possible value for \(\xcancel{m}\), it's not the only one.

Lemma. Given \(\xcancel{x^2 + mx + n = 0}\) and \(\xcancel{x = -3}\), there are countably infinite possible values for \(m\) such that \(\xcancel{m,n \in \mathbb{Z}}\).

Proof. Rewriting the original equation with \(\xcancel{x = -3}\), \(\xcancel{9 - 3m + n = 0 \leadsto m = \frac{n}{3} + 3}\). It follows that \(\xcancel{\forall m \in \mathbb{Z}, \exists n \in 3\mathbb{Z}}\) such that \(\xcancel{m = \frac{n}{3} + 3}\). Since \(\xcancel{3\mathbb{Z} \subset \mathbb{Z}}\), and there exists a bijection between \(\xcancel{\mathbb{N}}\) and \(\xcancel{3\mathbb{Z}}\), i.e. \(\xcancel{\{1,2,3,4,5,6,...\} \longleftrightarrow \{0,3,-3,6,-6,9,...\}}\), there are \(\xcancel{\aleph_{0}}\) possible values for \(\xcancel{m}\). \(\xcancel{\blacksquare}\)