<s>[INST] Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:
$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$
$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.
$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$
$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 - 2xy\sqrt{7}\right) = 1$ 
$(x^2 + 7y^2)^2 - 7(2xy)^2 = 1$ (This is of the form $X^2 - 7Y^2 = 1$)
I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired
$X^2 - 7Y^2 = 1$
I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.
$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$
And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 - 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting: 
$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$
But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:
$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$ 
which simplifying I got
$(x^2-7y^2) \left( (x^2 + 7y^2)^2 - 7(2xy)^2 \right)$
I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!
 [/INST] Consider
$$(8+3\sqrt7)^2=127+48\sqrt7,$$
$$(8+3\sqrt7)^3=(8+3\sqrt7)(127+48\sqrt7)=2024+765\sqrt7,$$
$$(8+3\sqrt7)^4=(8+3\sqrt7)(2024+765\sqrt7)=32257+12192\sqrt7$$
etc. Then the first few solutions are $(8,3)$, $(127,48)$, $(2024,765)$,
$(32257,12192)$ etc.
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