More generally, with coefficients $n$ and $n - 1$ we obtain
For example, when the initial coefficient pair is $(3,2)$ the sequence of powers gives coefficient pairs $(j, j - 1)$ that sum to $5^{m}$. That $3$ is odd corresponds to the fact that $5 = 1 \textrm{mod} 4$, since we write $n = 2k + 1$ and the sum is then $4k + 1$. The matrix sequence demonstrates the obvious fact that $5^m = 1 \textrm{mod} 4$ for all $m$, as the first coefficient is always odd.
A number theorist will tell you that an odd prime $p$ equal to $1 \textrm{mod} 4$ satisfies $p = a^2 + b^2$ for unique $a$ and $b$. The integer ratio $a/b$ comes from $(2k)!$. For $p = 5$, $a$ and $b$ are usually found from the Gaussian integer factorization $5 = (2 + i)(2 - i)$, where $a = 2$ and $b = 1$. Each Gaussian factor may be expressed as another basic two crossing braid, as in
where the factors also come from the Jordan product. The number $x$ (with $x^2 = 1$) is omitted here, since it always appears in the same places. Note that the square of one such factor multiplies (up to signs) the first coefficient by $3$ and the second by $2$, as in the matrix above. How nice to see products of basic $B_3$ configurations expressing simple arithmetic rules.
Note that the Gaussian factor braid differs only by one crossing flip from the braid in the last post, giving two overcrossings instead of one.
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