Integer points on a hyperbola Why I'm here
I have the following problem in probability from a book:
You have a bag with red and white balls and you draw two balls without replacing. If the probability of drawing 2 white balls is exactly 50%, how many balls are in the bag?
Suppose x is the number of red balls and y is the number of white balls. I can figure out the smallest solutions by trying out small numbers: $(1,3)$ and $(6,15)$. But are there any others?
My Work
The probability of getting two white balls (call that e) is
$\mathbb{P}(e)=\frac{1}{2}=\frac{y}{x+y}\cdot\frac{y-1}{x+y-1}$
Which gives me this quadratic equation:
$x^2+2xy-y^2-x+y=0$

And any positive integer points on this curve should be solutions to the problem. All I need to do is to find the integer points.
First I had the idea that if I draw a line with rational slope from the first point, I should reach a second rational point in the hyperbola. If I connect the two points I found, I get a line with equation $y=12/5(x-1)+3$,
and because there are no integer solutions between (1,3) and (6,15), I want a line with higher slope. On the other hand, I can tell that the hyperbola has an asymptote $y=(1+\sqrt2)x+1/2$, so I want a line with a slope
below $1+\sqrt2$, or the second intersection will be in the wrong branch of the hyperbola.
I had a vague idea that continued fractions help with diophantine equations, so I decided to try using the convergents of $1+\sqrt2$ to get my rational points. They're guaranteed to be above $12/5$ and exactly half of them
will be below $1+\sqrt2$, so it's worth a try. These are the points I came up with:
(only every other slope will give a positive point)
slope      | x      | y      | 
-----------+--------+--------+
1+5/7      | 6      | 15     |
1+12/17    | -35    | -84    |
1+29/41    | 204    | 493    |
1+70/99    | -1189  | -2870  |
1+169/239  | 6930   | 16731  |
1+408/577  | -40391 | -97512 |
1+985/1393 | 235416 | 568345 |

And here is a surprise: all the numbers so far are integers! They all also satisfy the initial question, and every other convergent of $\sqrt2$ is giving me a new solution to the problem. I have tried other lines with rational slopes
(by averaging the slopes of two consecutive solutions) but so far, I can't find any other integer solution.
My Question
Are all (sub-) convergents of $\sqrt2$ going to  give me an integer point in the positive part of the hyperbola?
Are there any other integer points in the positive part of the hyperbola?
I know that the general quadratic equation in integers was solved by Lagrange, but this method seems really different from what I'm doing here, it only uses the continued fraction to find the first solution
(so not all convergents produce a solution), and then produces a recursive function for the rest of them. Is there any relation here?
Also, if you'd be so kind, could you suggest some expository material around quadratic equations in integers?
Other Stack Exchange questions
The following questions have been helpful to me so far:

*

*How to solve inhomogeneous quadratic forms in integers?

*Convergents as solutions for Pell's equation
 A: 
I know that the general quadratic equation in integers was solved by
  Lagrange, but this method seems really different from what I'm doing
  here, it only uses the continued fraction to find the first solution
  (so not all convergents produce a solution), and then produces a
  recursive function for the rest of them.

I like Dario Alpern's online calculator.
Enter the coefficients and push "show steps".
It gives a nice explanation about how it solves the equation.
It seems to be the cited method:

$x^2 + 2 ⁢x⁢y - y^2 - x + y = 0$
The discriminant is $b^2− 4⁢a⁢c = 8$
Let $D$ be the discriminant. We apply the transformation of Legendre 
  $Dx = X + \alpha$, $Dy = Y + \beta$, and we obtain:
$\alpha = 2⁢c⁢d - b⁢e = 0$
$\beta = 2⁢a⁢e - b⁢d = 4$
$X^2 + 2 X⁢Y - 1 = -16 \quad (1)$
where the right hand side equals $−D (a⁢e^2− b⁢e⁢d + e⁢d^2 + f⁢D)$
We will have to solve several quadratic modular equations. To do this
  we have to factor the modulus and find the solution modulo the powers
  of the prime factors. Then we combine them by using the Chinese
  Remainder Theorem.
The different moduli are divisors of the right hand side, so we only
  have to factor it once.
$-16 = −2^4$
Searching for solutions $X$ and $Y$ coprime.
We have to solve: $T^2 + 2 T - 1 \equiv 0 \pmod {-16 = −2^4}$
There are no solutions modulo $2^4$, so the modular equation does not
  have any solution.
Let $X' = 2⁢X$ and $Y' = 2⁢Y$. Searching for solutions $X’$ and $Y'$ coprime.
From equation $(1)$ we obtain $X'^2 + 2 X'⁢Y' - 1 = -16 / 2^2 = -4$
We have to solve: $T^2 + 2 T - 1 \equiv 0 \pmod{-4 = −2^2}$
There are no solutions modulo $2^2$, so the modular equation does not
  have any solution.
Let $X' = 4⁢X$ and $Y' = 4⁢Y$. Searching for solutions $X'$ and $Y'$ coprime.
From equation $(1)$ we obtain $X'^2 + 2 X'⁢Y' - 1 = -16 / 4^2 = -1$
We have to solve: $T^2 + 2 T - 1 \equiv  0 \pmod{ -1 = −1}$
$T = 0$
The transformation $X' =  k \quad (2)$ converts $X'^2 + 2 X'⁢Y' - 1 = -1$ to $P⁢Y'^2 + Q⁢Y'k + R⁢ k^2 = 1\quad (3)$
    where: $P = (a⁢T^2 + b⁢T + c) / n = 1$, $Q = −(2⁢a⁢T + b) = -2$, $R = a⁢n = -1$
The continued fraction expansion of $(Q + \sqrt{D / 4}) / R = (-1 +\sqrt{ 2}) / 1$ is:
    $0+ // 2// \quad (4)$
Solution of $(3)$ found using the convergent $Y' / (−k) = 1 / 0$ of $(4)$
From $(2)$: $X' = 0$, $Y' = 1$
$X = 0$, $Y = 4$
$X + \alpha = 0$, $Y + \beta  = 8$
Dividing these numbers by $D = 8$:
$x = 0$
    $y = 1$
$X = 0$, $Y = -4$
$X + \alpha= 0$, $Y + \beta = 0$
Dividing these numbers by $D = 8$:
$x = 0$
    $y = 0$
The continued fraction expansion of $(−Q +\sqrt{ D / 4}) / (−R) = (1 + \sqrt{2}) / (-1)$ is:
    $-3+ // 1, 1, 2// \quad (5)$
Solution of $(3)$ found using the convergent $Y' / (−k) = 1 / -2$ of $(5)$
From $(2)$: $X' = -2$, $Y' = 1$
$X = -8$, $Y = 4$
$X + \alpha = -8$, $Y + \beta = 8$
Dividing these numbers by $D = 8$:
$x = -1$
    $y = 1$
$X = 8$, $Y = -4$
$X + \alpha = 8$, $Y + \beta = 0$
Dividing these numbers by $D = 8$:
$x = 1$
    $y = 0$
$\fbox{$\fbox{$x = 0 \\ y = 1$}$}$
$\fbox{$\fbox{$x = 0 \\ y = 0$}$}$
Recursive solutions:
$x_{n+1} =  x_n + 2 ⁢y_n - 1\\y_{n+1} = 2 ⁢x_n + 5 ⁢y_n - 2$
and also:  
$x_{n+1} = 5 ⁢x_n - 2 ⁢y_n + 1\\y_{n+1} = - 2 ⁢x_n + y_n$
Written by Dario Alpern. Last updated on 25 July 2018.

A: If you follow the procedure suggested in the accepted answer in your first link, the equation becomes $X^2-8Y^2=8,$ and then you can write $V=2Y$ to get $X^2-2V^2=8,$ so you do get all solutions by looking at convergents to $\sqrt2$.  Of course the procedure involves setting $Y=2x+2y-1$ and $X=8y-4,$ so it's not immediate that integral solutions for $X,V$ will lead to integral solutions of $x,y,$ but the reverse is clearly true.  Given your numerical results, I'd bet you always do get integral $x,y,$ and I'd expect it to be easy to show.
A: Sorry not an answer but too much for a comment. I was observing how a brute force C solution confirms the values you posted, but the negative ones are unusual.
My program tests for the condition $2W(W−1)=B(B−1)$ where $B=R+W$.
You posted these results 

Red     White
1       3
6       15
-35     -84
204     493
-1189   -2870
6930    16731
-40391  -97512
235416  568345

The results I got are 

Red     White
1       3
6       15
35      85
204     493
1189    2871
6930    16731
40391   97513
235416  568345

The interesting part is that where you have negative values, I have the same positive values, but white is 1 greater than your absolute value.
This is the code:
#include <stdio.h>

#define MAXB    568346

int main()
{
    long long B, R, W;

    for(R = 1; R <= MAXB; R++) {
        for(W = 1; W <= MAXB; W++) {
            B = R + W;
            if(2 * W * (W - 1) == B * (B - 1)) {
                printf("R=%lld W=%lld\n", R, W);
            }
        }
    }
}

