Square numbers in the form $1+4y$ 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?
 A: 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.
A: 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:
$$
\forall p \in \mathbb{N}, \quad x = p+1, \quad y = p(p+1)
$$
