(x^3/N) approximately= (x^2/y) (25/17) approximately= (85/58) Thoughts?
All my hypotheses in this thread rely on these 4 equations being true. p3 = ((pnp^2 + x^3) / pnp) – ((pnp + (x^2 / (pnp^2 +
x)) * pnp)) p5 = (pnp^2 + x^3) / pnp - (x^3 /
pnp) Separate equations. pnp=x*y x&y are Prime
factors of semi-Prime pnp Don’t simplify, graph in software.
x = Sqrt[ [ ((x^2 * pnp^4 + 2 * pnp^2 * x^5) + x^8) /
pnp^4]] [pnp^4 = [[ ( (pnp^4 * x^2 + 2 * pnp^2 *
x^5) )] / x^2 ] – [(x^3 / pnp / 2)]