Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program?

Question

I am reading a puzzle that ends up based on the given info to require the solution of the following equation:
$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$
also knowing that:
$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$ and $x >= 0 \land y >= 0$

The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).

I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?

Trying out I could reach nowhere to be honest:
$$\begin{align} 2x(x-1) &= (x + y) (x + y -1) \\[4pt] 2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt] x^2 - x &= y^2 +2xy -y \end{align}$$

but I don't see how this helps at all.

Is there any way to do some progression?

Answer 1

Let $z=x+y$, then $$2(2x-1)^2=(2z-1)^2+1$$ This is a Pellian equation, which you can look up.

Answer 2

This extends @Empy2's answer.

Starting from $2x(x-1) = (x+y)(x+y - 1) := z(z-1)$, we complete the squares:

$$2(x^2-x+\frac14)-\frac12 = z^2-z + \frac14 - \frac14$$

$$2(x-\frac12)^2 = (z-\frac12)^2 + \frac14$$

$$2(2x-1)^2 = (2z-1)^2+1$$

Equations of the form $x^2 - Dy^2 = \pm1$, where $D$ is not a square, are the simplest Pell equations (as opposed to the generalized version). Their solutions are related to the convergents of $\sqrt D$ (here, $\sqrt 2$.)

http://oeis.org/A000129 gives some details of these convergents, whose numerators and denominators give the solutions:

$$1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741\dots$$

These fractions have a pattern:

$$1^2-2\times 1^2 = -1$$

$$3^2-2\times 2^2 = +1$$

$$7^2 - 2\times 5^2 = -1$$

$$17^2-2\times 12^2 = +1$$

In particular: $$239^2 - 2\times169^2 = -1$$

which gives the solution $x=85, z = 120$, and is the only one in range ($199 < 2z - 1 < 999$).

Answer 3

If this of some help for this arithmetical problem, here is a vizualisation of the domain (right hand side quarter) defined by inequations

$$x>y \ \ \text{and} \ \ x+y>100$$

(constraint $x+y \le 500$ is not taken into account)

with a red dot for the proposed solution and (in black) the curve of the hyperbola defined by the equation. What you have to show is that it the unique point of the hyperbola with integer coordinates in the domain.

Answer 4

I am guesing $x,y$ are intergers and you have problems programming so here is my solution.

You have $x^2-x=y^2+2xy-y$ then $x^2-y^2-2xy=x-y \rightarrow (x-y)^2-2y^2=x-y$.

let $z=x-y$, an positive interger. You will have:

$z^2-z-2y^2=0$, so $z$ (since it is positive) must be: $z=\frac{1+\sqrt{1+8y^2}}{2}$.

Since $2y-z=x+y$ you have that $100 \leq 2y+ \frac{1+\sqrt{1+8y^2}}{2} \leq 500$.

Now find a function plotter online, or Geogebra and plot (lets reverse the roles of $x$ and $y$, so now $y$ is named $x$ and vice versa) $f(x)=2x+ \frac{1+\sqrt{1+8x^2}}{2} $

You can plot the lines $y=100$ and $y=500$ and find wich intergers get plotted between those lines. This shortens your search.

Also you can plot $y > x$ (remember we revrsed the roles) and check which intergers image falls in the intersection of $y\geq 100 \cap y \leq 500 \cap y\geq x$.

This will give you values for $x$ (or likely value). This will shorten or end your search.

Hope it helps.