Factorise This?

One way of generating infinite solutions is to use triangular numbers.

WLOG $x > y$ .

Let $x$ be $\frac{n(n+1)}{2}$ . Then make $y$ be $\frac{n(n-1)}{2}$ .

$x-y = \frac{2n}{2} = n$ .

$x+y = \frac{2n^2}{2} = n^2$ .

Therefore $(x-y)^2 = x+y$ , and there are an infinite amount of solutions.

This can also be represented geometrically:

As $x$ and $y$ are successive triangular numbers, they can be made into a square. However, instead of using a grid, we will use dots, with one dot in each grid square.

The area of the square is equal to $x + y$ , and the diagonal of the square is equal to $x-y$ , as removing it would leave two equal triangles, meaning it is the difference between $x$ and $y$ .

The number of dots along the diagonal of the square is equal to the number of dots along each of the edges, meaning that $x-y = \sqrt{x+y}$ , satisfying the equation.

A diagram for n = 3 can be seen below:

Put $x-y=k \implies x=y+k$ in the given equation

$k^2=2y+k \implies 2y=k(k-1) \implies y=\dfrac{k(k-1)}{2}$

Now $x=k+\dfrac{k(k-1)}{2}=\dfrac{k(k+1)}{2}$

We can find infinitely many $k$ .For each $k$ $(x,y)$ will be integer because $k(k-1)/2$ and $k(k+1)/2$ are integers as thay are product of consecutive integers.These evens are divided by $2$ so they will be integers.

So, there will be infinitely many solutions.

It is given that $(x-y)² = x + y$ .

We note that $( x-y)² = (x+y)² -4xy$ , then we have

$(x+y)² - (x-y)² = 4xy \Rightarrow (x+y+x-y)(x+y-x+y) = 4xy$

$4xy = 4xy$ , then, we can conclue that there are infinte pairs of $(x,y)$ .