Let me make a tiny-microscopic improvement. Let   $n := n(x)$.   Then

$$ P(x)\ |\ (n-1)\cdot(n-2)\ =\ (n-\frac 32)^2 - \frac 14$$

It follows that:

THEOREM

$$ n(x)\quad \ge\quad \left\lceil\sqrt{P(x)+\frac 14}\ +\ \frac 32\right\rceil $$

Let me make a tiny-microscopic improvement. Let   $n := n(x)$.   Then

$$ P(x)\ |\ (n-1)\cdot(n-2)\ =\ \left(n-\frac 32\right)^2 - \frac 14$$

It follows that:

THEOREM

$$ n(x)\quad \ge\quad \left\lceil\sqrt{P(x)+\frac 14}\ +\ \frac 32\right\rceil $$

Let me make a tiny-microscopic improvement. Let   $n := n(x)$.   Then

$$ P(x)\ |\ (n-1)\cdot(n-2)\ =\ (n-\frac 32)^2 - \frac 14$$

It follows that:

THEOREM

$$ n(x)\quad \ge\quad \left\lceil\sqrt{P(x)+\frac 14} + \frac 32\right\rceil $$

Let me make a tiny-microscopic improvement. Let   $n := n(x)$.   Then

$$ P(x)\ |\ (n-1)\cdot(n-2)\ =\ (n-\frac 32)^2 - \frac 14$$

It follows that:

THEOREM

$$ n(x)\quad \ge\quad \left\lceil\sqrt{P(x)+\frac 14}\ +\ \frac 32\right\rceil $$