Find all natural numbers $n$ for which the equation $x(x+n)=y^2$ does not have any solutions over the positive integers I tried rearranging it and factoring the sum of squares, so that I get
$$xn=(y-x)(y+x)$$
But at this point I have just no clue how to continue. I tried to manipulate the fact that $n$ divides right hand side, but it didn't work at all, so I don't even consider that attempt worth putting there. I also tried playing around with parity and checking the cases, but it also was pointless.
So, please, help me with that. I've been trying to solve this problem for the eternity
 A: Where both $x$ and $y$ are required to be positive, this has a solution for each positive $n \ge 3$ with $n$ odd or $n \ge 6$ with $n$ even: Indeed, $x(x+n)$ is a square if both $x$ and $x+n$ is a square. We use this repeatedly.
Then for each odd integer $n \ge 3$ there is an $x,y \in \mathbb{Z}^+$ such that $x(x+n)$ is a square.  Indeed, write $n=2j+1$. Then set $x=j^2$. Then $x+n=(j+1)^2$, so $(j^2)(j^2+2j+1)=x(x+n)$ is a square.
For each integer $n \ge 6$; $n \equiv_4 2$ there is an $x,y \in \mathbb{Z}^+$  such that $x(x+n)$ is a square. Indeed, let $x'$ be such that $x'\left(x'+\frac{n}{2}\right)$ is a square; as $\frac{n}{2}$ is odd there is such an $x'$ by the above paragraph. Then setting $2x'=x$, then $x\left(x+n\right) =$ $4x'\left(x'+\frac{n}{2}\right)$ is a square.
For each integer $n \ge 8$; $n \equiv_4 0$ there is $x,y \in \mathbb{Z}^+$  such that $x(x+n)$ is a square. Indeed, $n$ can be written $n=(2j+1)+(2j+3)$ for some nonnegative integer $j$; then setting $x=j^2$ note that $x+n=(j+2)^2$ so $x(x+n)$ is a square.
For $n=1,2$ this has no solutions with $x$ and $y$ both positive; indeed,
if $x$ is odd or $n=1$ then $x$ and $x+n$ are relatively prime and so both $x$ and $x+n$ must be squares and this is clearly impossible as $ |a^2-b^2|$ is at least 3 for any 2 distinct positive integers $a$ and $b$ and at least 5 if $a \ge 3$; if $n=2$ and $x$ is even then $\frac{x}{2}\left(\frac{x}{2}+1\right)$ is a square; $\frac{x}{2}, \left(\frac{x}{2}+1\right)$ integers. Can you finish from here.
EDIT: A similar line of reasoning can be shown that this has no solutions with $n=4$ and $x,y$ both positive.
A: We can use the discriminant: $ax^2 + bx + c = 0$ has rational solutions iff $D = b^2 - 4ac$ is a perfect square. Let's put $x(x+n) = y^2$ in the form $ax^2 + bx + c = 0$: $$x^2 + nx - y^2 = 0 \text{ has } a = 1, b = n, c = -y^2,$$ therefore integer solutions are only possible when $D = n^2 - 4(1)(-y^2) = n^2 + 4y^2$ is a perfect square.
In such a case, $(n, 2y, \sqrt{D})$ form a Pythagorean triple, and from looking at the parametrization for Pythagorean triples, it is clear that any natural number $n$ other than $1$, $2$, or $4$ can be in a Pythagorean triple $(n, 2y, \sqrt{D})$ for some nonzero integer $y$.
(If $n = 1$ or $n = 2$, then the only possibility for $2y$ is $2y = 0$. $n = 4$ doesn't work because in that case the only nonzero possibility for $2y$ equals $3$.)
Therefore the equation $x(x+n) = y^2$ has positive integer solutions $(x, y)$ for all natural numbers $n$ other than $1, 2, 4$. $\blacksquare$
