I want to solve the equation $y+x=x^2$: $$ x^2-x-y=0 \\ x_{1;2}=\frac{1\pm \sqrt{1+4y}}{2} $$
However I want the solutions to be only natural numbers; the question then turns to find values of $y$ such that $1+4y$ is always a square number. I have no idea how to solve this problem.
Could anyone help me?
On
If you want $x$ and $y$ to be natural number then you correctly found that $1+4y$ must be a square.
So there exists $z \in \mathbb{N}$ such that:
$$ z^2 = 1+4y $$
It is pretty obvious that $z$ is odd. So you can rewrite $z$ as:
$$ \exists p \in \mathbb{N}, \quad z = 2p+1 $$
Then you substitute and get:
$$ (2p+1)^2 = 1+4y \\ \Leftrightarrow 1+4p(p+1) = 1+4y $$
You can identify: $y = p(p+1)$ This is a necessary condition for the equation to have a solution.
Then we check that if $p \in \mathbb{N}$, and $y = p(p+1)$ satisfies the equation. According to your own finding we then have:
$$ x_1 = (1+z)/2 = (1+2p+1)/2 = p+1 \\ x_2 = (1-z)/2 = (1-2p-1)/2 = -p $$
As we want $x \in \mathbb{N}$ we cannot have $x < 0$. So the solution are:
$$ \forall p \in \mathbb{N}, \quad x = p+1, \quad y = p(p+1) $$
If $1+4y$ is a square, it means that there exists $z\in \Bbb N$ such that $$z^2=1+4y.$$ Therefore, solving for $y$, $$y=\frac{z^2-1}{4},$$ and that is the expression of the $y$'s you are looking for.