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I know that Pell's equation $x^2-dy^2=1$ always has solutions and I want using that fact show that $x^2-dy^2=4$ also always has solutions.

$$ x^2-dy^2=4\tag{*} $$

I try something like this... Let $(x_1,y_1)$ be solution of $x^2-dy=4$. Lets look all cases that can be dependent on odd or even number $x$ or$y$. It is clear that can't be je $x_1$ even, and $y_1$ odd, so we have only 3 cases.

If $x_1, y_1$ both odd, then by dividing equation by 4 I get the number 1 on the right and according to Theorem the equation has a solution.

If $x_1$ is odd, and $y_1$ even, then $4 \mid d$, tj. $d=4d'$. Then again by dividing equation by 4 I get the number 1 on the right and according to Theorem the equation has a solution.

How to finish this proof?

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    $\begingroup$ Do you want to show it has a solution or to find all solutions? $\endgroup$
    – Bernard
    Jul 16, 2020 at 9:57
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    $\begingroup$ If $(x,y)=(a,b)$ is a solution to $x^2-dy^2=1$, then $(x,y)=(2a,2b)$ is a solution to $x^2-dy^2=4$. (Not all solutions to $x^2-dy^2=4$ can be obtained this way. For example, $x^2-5y^2=4$ has a solution $(x,y)=(3,1)$.) $\endgroup$ Jul 16, 2020 at 10:06
  • $\begingroup$ @Bernard just show that has a solution $\endgroup$
    – josf
    Jul 16, 2020 at 10:09
  • $\begingroup$ @josf This has been shown by Batominovski. $\endgroup$
    – Peter
    Jul 16, 2020 at 11:34
  • $\begingroup$ @josf this will always have a solution but not necessarily a primitive solution. But it can be proved under what conditions, it will have a primitive solution. For example, batominovski showed a primitive solution for d = 5. $\endgroup$
    – Math Lover
    Jul 16, 2020 at 13:08

2 Answers 2

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I know that Pell's equation $x^2-dy^2=1$ always has solutions and I want using that fact show that $x^2-dy^2=4$ also always has solutions.

As Batominovski indicated in comments, if you know a solution $(x,y)$ to $x^2-dy^2=1$,

then $(X,Y)=(2x,2y)$ is a solution to $X^2-dY^2=4$,

because $X^2-dY^2=(2x)^2-d(2y)^2=4x^2-d4y^2=4(x^2-dy^2)=4(1)=4.$

For example, solutions to $x^2-5y^2=1$ are $(x,y)=(1,0), (9,4), (161,72), ...$,

so solutions to $X^2-5Y^2=4$ are $(X,Y)=(2,0), (18,8), (322, 144), ...$.

So this shows that $X^2-5Y^2=4$ has solutions, though there are solutions this does not find,

such as $(X,Y)=(3,1),(7,3),(47,21),(123,55),(843,377), ...$.

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$$x^2-dy^2=4\implies d=\frac{x^2-4}{y^2}\quad \lor\quad y=\sqrt{\frac{x^2-4}{d}}$$

The latter seems easier to understand. As long as $(x^2-4)$ is multiple $(d)$ of a perfect square, there is a solution. Try these $(x,d)$ pairs and you will see they yield a natural number for $y$. $$(4,3)\quad (6,2)\quad (7,5)\quad (8,15)\quad (10,6)\quad (14,3)\quad ...\quad (23,1)\quad ...$$ You can do find these yourself in a spreadsheet as I did.

Set $x=2$ in a spreadsheet and have every row/cell under it increment by $1$. In the cell to the right of $x=2)$ enter the formula for $x^2-4$ and fill down. You will then be able to see which "squares-minus-4" are squares when divided by a mentally calculated $d$.

There are "repeating patterns" of a sort, for instance, $10\rightarrow96$ and $14\rightarrow192$ so once you know that $96/6=16$, you can see that $192$ is a multiple of $95$ and must have a solution, in this case $192/3=64$. You may have to experiment a little bit or even write a little program to find such as $(22,19)\rightarrow25$ but you will always find solutions.

I cannot help you with "proof" but, you may be able to figure something out from the "several" repeating patterns I saw in MY spreadsheet.

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