# 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 $$\xcancel{m}$$. $$\xcancel{\blacksquare}$$