This article presents a method of solving the Pell equation
$x^{2}-A y^{2}=1$ (where $A$ is a non-square integer) through a specific formulation.

Proposition For the equation (1), express $A$ in the form $A=\left(\frac{p}{q}\right)^{2}+B$ (where $p$ and $q$ are positive integers, $p>q, (p, q)=1, B \neq 0$), such that $\left(A q^{2}-p^{2}\right) \mid 2 p$. Then,
i) When $B>0$, a positive integer solution to (1) is
$$
y=\frac{2 p q}{A q^{2}-p^{2}}, x=\frac{2 p^{2}}{A q^{2}-p^{2}}+1 \text {; }
$$
ii) When $B<0$, a positive integer solution to (1) is
$$
y=\frac{2 p q}{p^{2}-A q^{2}}, x=\frac{2 p^{2}}{p^{2}-A q^{2}}-1 .
$$

Example If $A=(2 h m)^{2} \pm 8 h, m, h \in N$, find the solution to equation (1).
Solution: Let $A=\left(\frac{p}{q}\right)^{2}+B$, then $p=2 h m^{2} \pm 2$, $q=m$. Therefore,
$$
B=A-\left(\frac{p}{q}\right)^{2}=-\frac{4}{m^{2}}<0
$$

and $\frac{2 p}{A q^{2}-p^{2}}=-h m^{2} \mp 1$,
which means $\left(A q^{2}-p^{2}\right) \mid 2 p$.
Thus, the solution to (1) is
$$
\left\{\begin{array}{l}
y=h m^{3} \pm m, \\
x=2\left(h m^{2} \pm 1\right)^{2}-1
\end{array}\right.
$$