Consider a circle of circumference \(4C^z\),
The circumference along arc EDF is of length \(C^z\).
Let \(p\) be a factor of \(C^z\), we draw another smaller circle of circumference \(\cfrac{4C^z}{p}\). The quadrant arc is of length \(\cfrac{C^z}{p}\). This arc is of whole integer length as \(p\) is a factor of \(C^z\). And this quadrant arc scales to \(C^z\) by \(p\).
A line OD is drawn to divide the smaller circle such that,
\(m+n=\cfrac{C^z}{p}\)
where both \(m\) and \(n\) and \(\cfrac{C^z}{p}\) are integers. Markings on the longer arc EDF is of integer width being divided by \(C^z\). Markings on arc GHI is also of integer width being divided by \(\frac{C^2}{p}\). The shorter arc GHI is scaled by \(p\) to the longer EDF. Line OD from O intersect a integer marking on arc GHI and then intersect another integer marking on the arc EDF. The width of the markings on arc EDF is \(p\) times smaller than the width on arc GHI. One marking on GHI scale to \(p\) markings on EDF.
This line divides \(C^z\) into two integers, \(K\) and \(J\).
As the circumference \(C^z\) was reduced by \(p\), The arc \(K\) is,
\(K=mp\)
and arc \(J\) is,
\(J=np\)
Both K and J have a factor \(p\) and are integers. So,
\(K+J=(m+n)p=\cfrac{C^z}{p}p=C^z\)
and all \(K\), \(J\) and \(C^z\) has a common factor \(p\).
Suppose \(p=ab\) and we scale to \(a\) now,
arc \(G'H'I'=\cfrac{C^z}{a}=\cfrac{C^z}{p}b\). When we divide arc G'H'I' into equally space markings, their width is,
width of markings on arc G'H'I\(=\cfrac{p}{bC^z}=\cfrac{1}{b}\times\)width of markings on arc GHI.
The marking has scaled proportionately to accommodate line OD. So, when there is a factor of \(C^z\), all other factors of \(C^z\) are also factors of \(K\) and \(J\).
If \(p\) is prime, that \(p=ab\), either \(a=1\) and \(b=p\), or \(a=p\) and \(b=1\), and \(pa=b\), either \(a=1\), \(p=b\) or \(b=1\) and there is no scaling, the arc remains on \(C^z\). In this case of prime \(p\), as the only (repeated) factor of \(C^z\), all markings along quadrant arcs above and below \(p\) will not accommodate line OD. None of the markings generated above and below \(p\) will intersect the line OD, because arc GHI drawn are not of integer length.
Both \(K\) and \(J\) share this prime factor.
This does not prove Beal Conjecture, but if given a plot of actual \(A^x\), \(B^y\) on a quadrant arc of length \(C^z\), and a line OD through center O dividing\(C^z\) into \(A^x\) and \(B^y\), any quadrant arc that intersects OD at integer markings along the arc are factors of both \(A^x\) and \(B^y\) and so, also a factor of \(C^z=A^x+B^y\).
As for the conjecture, if there is a factor of \(C^z\), then both \(A^x\) and \(B^y\) share this factor. Other factors that divide \(C^z\) into integers are also possible on the same plot. These factor cannot be prime. If \(C^z\) has one prime factor then no other factor are possible on the plot. Both \(A^x\) and \(B^y\) also have this prime factor.
