# Strange Mathematics

Eric Bailey

Posted on 2 August, 2020

Tags: mathematics, proof, comics

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

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?

SIX.

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 exapnds 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 \(\xcancel{m}\).~~
~~\(\xcancel{\blacksquare}\)~~