Erdős–Straus Conjecture In Odd Steps

 This is the reason why a half step was take,

a old number of steps is required to reach length $1$ at $\cfrac{\pi}{2}$ further down.  Without the half steps the formulation of $L$ will scream for an even factor.

$M=4\cfrac{N}{n}$ is an even partition such that integer steps can be taken to $N$.  $a$ is odd because $N$ is odd due to $n$ being odd.  $b$ is also odd for the same reason.  So $c$ is even such that $a+b+c$ fills up $M$ an even number.  But full steps in $c$ cannot reach $\cfrac{\pi}{2}$.  $L$ is in odd steps only due to $N$ and $n$ being odd.  So, a half step is taken to find the step count and $c$ is recovered by doubling the half step count ( times $2$).

It is possible instead to step all the way to $\cfrac{3\pi}{2}$ and then divide the count obtained by $3$?  In this case the expression for length $\bar{c}$ is$\cfrac{3}{L}$ which is the reciprocal of an integer only if $L$ has a factor $3$.  This is not the case for prime numbers other then $3$, and the prove will fail.

We start not with $n$ dividing a circle, we start with $4$ dividing the circle. 

Goodnight.