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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!

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  • $\begingroup$ $(8-3\sqrt7)(127-48\sqrt7)=2024-765\sqrt7$ $\endgroup$ – J. W. Tanner Apr 24 at 2:27
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    $\begingroup$ Since $(8x+21y)^2 - 7(3x+8y)^2 = x^2-7y^2$, if $(x,y)$ is a solution, so does $(8x+21y,3x+8y)$. $\endgroup$ – achille hui Apr 24 at 2:49
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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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  • $\begingroup$ Thank you! This approach makes a lot of sense, and is definitely much less of a headache than dealing with the extended variable mountain I was working towards. $\endgroup$ – Drew Pesall Apr 24 at 4:07
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Remember that $x^2-7y^2=1$, so you can write $x^6-21x^4y^2=-14x^4y^2+7x^4$ and keep going $$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1\\ -14x^4y^2+7x^4+147x^2y^4-343y^6=1\\ 7x^4+49x^2y^4-14x^2y^2-343y^6=1\\ 7x^4-14x^2y^2+49y^4=1$$ Now you can plug in your fourth degree solution and be done. You can also use the Brahmgupta-Fermat identity to reach the same result

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Beginning with pairs $(1,0)$ and then $(8,3),$ all the other soltuions with positive variables satisfy $$ x_{n+2} = 16 x_{n+1} - x_n \; , \; $$ $$ y_{n+2} = 16 y_{n+1} - y_n \; . \; $$ The $x_n$ begin $$ 1, 8, 127, 2024, 32257, 514088, 8193151, $$ The $y_n$ begin $$ 0, 3, 48, 765, 12192, 194307, 3096720, $$

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