Get ready Part 14

I'm not sure if I'm missing the idea behind the problem. Not sure why the entries are square, noting that the differences between different pairs of squares are not nescessarily unique (which can lead to more commuting matrices**). Anyhow, the set size seems higher than necessary:

In order to prove the statement we must find the largest set that does not contain pairwise commuting elements.: First note that there are $15^4$ possibilities.

Observing matrices of the form $\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ , commute together (and when $a=b$ , it commutes with all other matrices). Hence there can only be 1 of these, so we can discard $15^2 - 1$

Observing matrices of the form $\begin{pmatrix} a & x \\ 0 & a \end{pmatrix}$ , where $x$ is not $0$ , commute together. There are $15^2 - 15$ of these, so we can discard $15^2 - 15 - 1$ .

Simialrly for matrices of the form $\begin{pmatrix} a & 0 \\ x & a \end{pmatrix}$ .

Hence, currently, any set greater than $15^4 - 3*15^2 + 2*15 +3 = 49983$ contains a pair of commuting matrices.

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$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\begin{pmatrix} e & f \\ g & h \end{pmatrix}$ commute iff $bg = cf, (a-d)g = (e-h)c$ and $(a-d)f=(e-h)b$

For $b,c,f,g \neq 0$ we get that \begin{pmatrix} a & b \ c & d \end{pmatrix}) commutes with $\begin{pmatrix} y & bx \\ cx & y-x(a-d) \end{pmatrix}$ for values of $x,y$ that lead to valid entries.

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For $x = 1$ : $\begin{pmatrix} a_i & b \\ c & d_i \end{pmatrix}$ . These commute when $a_1 - d_1 = a_2 - d_2$ .

eg $\begin{pmatrix} 16 & 4 \\ 9 & 1 \end{pmatrix}$ and $\begin{pmatrix} 64 & 4 \\ 9 & 49 \end{pmatrix}$ commute.

Noting pairs of pairs such that $a_1 - d_1 = a_2 - d_2$ , we have: (9,0), (25,16)

(16,1), (64,49)

(16,0), (25,9)

(25,4), (121,100)

(25,1), (49,25)

(25,0), (169,144)

(36,9), (196,169)

(36,4), (81,49)

(49,9), (121,81)

(49,4), (81,36)

(49,1), (169,121), (64,16)

13 pairs. With $14^2 = 15^2 - 2*15 +1$ possible combinations of $b$ and $c$ . So we have to get rid of at least 12 pairs from each combination.

So an upper bound on the maximum set with no pairwise commuting matrices is $15^4 - 15^3 + 26*15 -9 = 47631$ .

Something probably went wrong in some places, and I'm sure I'm missing a lot, eg $A$ and $n^2 A$ commute.

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