Pell's equation $x^2-dy^2=4$ always has solutions 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?
 A: 
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 $(X,Y)=(3,1),(7,3),(47,21),(123,55),(843,377), ...$.
A: $$x^2-dy^2=4\implies d=\frac{x^2-4}{y^2}\quad \lor\quad y=\sqrt{\frac{x^2-4}{d}}$$
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.
