Prime Arrows and Targets

 From previously, "Going Around In Circle" dated 09 Jan 2023, $3$ has been excluded as a factor of $q+2$,

the total number of integer marks along arc $q+2$ from $1$ to $\cfrac{q+2}{3}$ is about,

$T=\cfrac{q+2}{3}-1+1$

after removing $1$ and $2$ as possible intercepts,

$T=\cfrac{q+2}{3}-3+1=\cfrac{q+2}{3}-2=\cfrac{q-4}{3}$

and the number of possible 1st intercept lines from arc $q$ and below is,

$a=\cfrac{q-1}{2}-2=\cfrac{q-5}{2}$

where arcs $1$ and $2$ are excluded.  There is more targets than arrows, and the numbers $2$, $3$, $4$ and $5$ are involved.

The targets are evenly spaced along arc $q+2$ at equal angle intervals, $\cfrac{\pi}{2}*\cfrac{1}{2}*\cfrac{1}{q+2}=\cfrac{\pi}{4(q+2)}$,

$\cfrac{3\pi}{4(q+2)}$,$\cfrac{4\pi}{4(q+2)}$, $\cfrac{5\pi}{4(q+2)}$...  $\cfrac{(q+2)\pi}{3*4(q+2)}=\cfrac{\pi}{12}$ excluded. 

$\cfrac{\pi}{4}\cfrac{3}{(q+2)}$,$\cfrac{\pi}{4}\cfrac{4}{(q+2)}$, $\cfrac{\pi}{4}\cfrac{5}{(q+2)}$... $\cfrac{\pi}{4}\cfrac{(q+1)}{3*(q+2)}$

whereas the arrows are at,

$(\cfrac{\pi}{4}*\cfrac{1}{3})$ excluded, $(\cfrac{\pi}{4}*\cfrac{1}{5})$,  $(\cfrac{\pi}{4}*\cfrac{1}{7})$, $(\cfrac{\pi}{4}*\cfrac{1}{9})$... $(\cfrac{\pi}{4}*\cfrac{1}{q-2})$

Removing $\cfrac{\pi}{4}$ the common factor,

$\cfrac{3}{(q+2)}$,$\cfrac{4}{(q+2)}$, $\cfrac{5}{(q+2)}$... $\cfrac{(q+1)}{3}*\cfrac{1}{(q+2)}$

$\cfrac{1}{5}$,  $\cfrac{1}{7}$, $\cfrac{1}{9}$... $\cfrac{1}{q-2}$

If these two series share a common member, then $q+2$ is not prime.

This is nothing new, as  a prime candidate is often tested for all possible factors.  These members are on the delimited line,

$\cfrac{1}{q+2} x$ 

and  $3\le x\le \cfrac{q+1}{3}$ 

and the curve, 

$\cfrac{1}{x}$ 

but for $x$ odd, $5\le x\le (q-2)$

If the line and the curve intersect within the specified domains at integer points then $q+2$ is not prime.

The target are on a line and the arrows on a reciprocal curve.  They will meet once, but do they meet in the restricted domains?  The line maybe translated along the $x$ axis, as any coincidental member on the two series will fail $q+2$ as a prime.

The line and curve will meet, but not necessarily on odd integer points, but always, no matter how large $q$ is, as the curve is translated along the $x$ axis within its domain in odd integer steps,$5\le x\le (q-2)$.

Integers that are candidates for factors will be below the curve $\cfrac{1}{x}$ when the line starts at the origin.  As the line translated along $x$ in odd integer steps, candidates  that are not factors are moved above the curve.

Factors will meet on the curve exactly.

Good morning.