写出 $\left(px - \dfrac{q}{x}\right)^n$ 的二项式展开式中的通项，其中 $p$ 和 $q$ 是正整数。 如果 $\left(px - \dfrac{q}{x}\right)^n$ 的二项式展开式中的第五项与 $x$ 无关，证明 $n = 8$。 由此，若已知第五项为 1120 且 $p - q = 1$，求 $p$ 和 $q$ 的值。

1. Write down the general term using the binomial theorem formula

The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $\binom{n}{r} a^{n-r} b^r$.
Substituting $a = px$ and $b = -\frac{q}{x}$:
$T_{r+1} = \binom{n}{r} (px)^{n-r} \left(-\frac{q}{x}\right)^r$

1. Substitute $r = 4$ into the general term to find the fifth term

The general term is $T_{r+1} = \binom{n}{r} (px)^{n-r} \left(-\frac{q}{x}\right)^r$.
For the fifth term, $r = 4$:
$T_5 = \binom{n}{4} (px)^{n-4} \left(-\frac{q}{x}\right)^4$
$T_5 = \binom{n}{4} p^{n-4} q^4 x^{n-4} x^{-4}$
$T_5 = \binom{n}{4} p^{n-4} q^4 x^{n-8}$

2. Equate the power of $x$ to zero to show $n = 8$

Since the term is independent of $x$, the power of $x$ must be $0$.
$n - 8 = 0$
$n = 8$ (shown)

1. Express the fifth term using $n=8$ and $r=4$

The fifth term is $T_5 = T_{4+1}$. Using $n=8$ from part (ii):
$T_5 = \binom{8}{4} (px)^{8-4} \left(-\frac{q}{x}\right)^4$
$T_5 = 70 (p^4 x^4) \left(\frac{q^4}{x^4}\right) = 70 p^4 q^4$

2. Equate the fifth term to 1120 and simplify

$70 p^4 q^4 = 1120$
$p^4 q^4 = 16$
$(pq)^4 = 16$
Since $p$ and $q$ are positive integers, $pq = 2$.

3. Solve the simultaneous equations for $p$ and $q$

We are given $p - q = 1$, so $p = q + 1$.
Substitute $p = q + 1$ into $pq = 2$:
$(q + 1)q = 2$
$q^2 + q - 2 = 0$
$(q + 2)(q - 1) = 0$
Since $q$ is a positive integer, $q = 1$.

4. Find the value of $p$

$p = 1 + 1 = 2$