The equation $x^2-dy^2=1$ will be transformed into a quadratic equation with the use of triangular numbers. The motivation to use the triangular numbers to express the squares $x^2$ and $y^2$ comes from the fact that the use of triangular numbers was successful in finding the sum of two and three squares of an integer $N$.
We set $x^2=T_n + T_{n+1}$ and $y^2=T_{n+k+1} + T_{n+k+2}$. The parameter $k$ is needed for situations where $x^2$ and $y^2$ are not consecutive squares. Then the Pell's equation becomes:
$$T_n + T_{n+1} -d*[T_{n+k+1} + T_{n+k+2}]=1$$
The final quadratic equation in $n$ is:
$$(d-1)*n^2 + [(4+2k)d-2]*n +d*(k^2 + 4k +4)=0$$
We consider the following example: $x^2-7y^2=1$. The quadratic equation in $n$ becomes: $$6n^2 + (26+14k)*n + 7k^2 + 28k +28=0$$
The discriminant $\Delta=4*(7k^2 + 14k + 1)$. We want $7k^2+14k+1=s^2$. This expression is equal to $13^2$ for $k=-6$. For the type of Pell's equation considered above, the value of $k$ is always negative since in this case the square $y^2$ is always smaller than $x^2$. Therefore the indices of the triangular numbers for $y^2$ will be smaller.
The integer value of $n$ is $n=7$. Now we can build the values of $x^2$ and $y^2$.
$$x^2=T_7+ T_8=8^2$$ and $$y^2=T_2 + T_3=3^2$$
so that $$8^2 -7*3^2=1$$
This method can easily be extended to the case of $x^2-dy^2=-1$ or to the case when the right hand side is different fro $1$ or $-1$. The form of the discriminant $\Delta$ seems to be $\Delta=4*(dk^2 + 2dk +1)$ for every example considered. This expression of $\Delta$ has now been derived.
Does this elementary method shed any light on the Pell's equation?