Find the sample size for a two-sided 90% CI when no prior estimate of $\hat{p}$ is given

Question

The following is not a homework problem, I'm preparing for my exam.

I worked to derive the n from Wilson's CI and I assumed that $\hat p = 0$ since there are no prior estimates but I got a negative value which (obviously) is wrong.

Answer 1

Let the unestimated $\hat p = \frac{x}{N}$

$1-\hat p = \frac{N-x}{N}$

$\sqrt{\frac{x(N-x)}{N^3}} = SEM = \frac{width}{2\times z_{alpha}}= \frac{0.2}{2\times1.645}$

http://www.sample-size.net/confidence-interval-proportion/

Expanding the above

we get $ x^2-Nx+0.0037N^3 = 0$

We know that $x$ has to be finite and positive so the determinant of the quadratic equation has to be positive

Thus $N^2 - 4\times 0.0037N^3 \ge 0$

$N\le67$ and size necessary is thus equal to $65$