Finding integers solutions for x² - y² = x + 2xy + y, and then proving some properties they seem to have

Question

I've found about 19 integer solutions, but they follow a very strange pattern.

x = 6, y = 2

35, 14

204, 84

1189, 492

Notice that every one of them equals the previous one times (3 + 2√2), and then rounded up. This is also true for every other integer solution I've found.

The ratio of x/y of all integer solutions approaches 2√2.

Note that the curve is not linear.

Answer 1

Short version: call the pairs $(x_n, y_n),$ and number $(x_0, y_0)= (0,0) \; , \; \; $ $(x_1, y_1)= (6,2) \; , \; \; $ $(x_2, y_2)= (35,14) \; , \; \; $ $(x_3, y_3)= (204,84) \; , \; \; $ we get rules (for my $n \geq 1$ ) $$ \color{red}{ x_{n+2} = 6 x_{n+1} - x_n \; , } $$ $$ \color{magenta}{ y_{n+2} = 6y_{n+1} - y_n + 2 \; .} $$

We could then solve for $x_n$ as $A (3 + \sqrt 8)^n + B (3 - \sqrt 8)^n $ by

In order to calculate $y,$ take instead $ z_n= y_n + \frac{1}{2},$ you may now solve for $z_n$ as $C (3 + \sqrt 8)^n + D (3 - \sqrt 8)^n$ because $ z_{n+2} = 6 z_{n+1} - z_n \; .$

ORIGINAL:

Here are the first few solutions to $w^2 - 2 v^2 = -1$ with both $w,v >0.$

See how the sequence $w_n$ obeys $w_{n+2} = 6 w_{n+1} - w_n, $ while $v_n$ $v_{n+2} = 6 v_{n+1} - v_n. $ We always get these Fibonacci type sequences, in our case one pair suffices.

Next, the thing you are setting equal to zero (times 4) comes out to $$ (2x-2y-1)^2 - 2(2y+1)^2 + 1, $$ so the desired equation comes out $$ \color{red}{ (2x-2y-1)^2 - 2(2y+1)^2 = -1. } $$

The annoying thing is that we need to allow four types of $x,y$ pairs (from positive $w,v$ from four choices $$ 2x-2y + 1 = w \; , \; \; \; 2y+1 = v \; , \; $$ $$ 2x-2y + 1 = w \; , \; \; \; 2y+1 = -v \; , \; $$ $$ 2x-2y + 1 = -w \; , \; \; \; 2y+1 = v \; , \; $$ $$ 2x-2y + 1 = -w \; , \; \; \; 2y+1 = -v \; . \; $$ For each of these we may solve for $y$ in terms of $v,$ then plug back in to solve for $x$ in terms of both $w,v.$

From your question, it appears that you prefer $ x > y > 0,$ that is going to mean the first choice, so we are getting $$ x = \frac{w+v }{2} \; , \; \; y = \frac{v-1}{2} $$

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 w^2 - 2 v^2 = -1 =   -1 *  1 

1,  7,  41,  239,  1393,  8119,  47321,  275807,  1607521,  

1,  5,  29,  169,  985,  5741,  33461,  195025,  1136689,  

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Answer 2

Your equation is $x^2-(2y+1)x-(y+y^2)=0$.

Solving this quadratic, $x=\dfrac{2y+1\pm\sqrt{(2y+1)^2+4(y+y^2)}}2=\dfrac{2y+1\pm\sqrt{8y^2+8y+1}}2.$

You can already see that $\dfrac xy=\dfrac{2+\frac1y\pm\sqrt{8+\frac8y+\frac1{y^2}}}2$ approaches $1+\sqrt2$ as $y\to\infty$.

In order for this $x$ to be an integer, $z=\sqrt{8y^2+8y+1}=\sqrt{2(2y+1)^2-1}$

must be an odd integer, and we must have $z^2-2Y^2=-1$, where $Y=2y+1$.

This is the well-known negative Pell equation, whose solutions are as follows:

$z=1,7,41,239,1393\dots$ and $Y=1,5,29,169,985\dots$, so we have $y=0,2,14,84,492\dots$.

The On-Line Encyclopedia of Integer Sequences gives the closed formula

$y_n = \dfrac{-4+(2+\sqrt2)(3+2\sqrt2)^n + (2-\sqrt2)(3-2\sqrt2)^n}8$,

and since $(2-\sqrt2)(3-2\sqrt2)^n<1$, I think from here you can see

that the next term is the previous term times $3+2\sqrt2$, rounded up.