(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)]