Diophantine equation $4x^2+n(n+1)x=y^2$ for a given $n>2$. I want to show that the equation $$4x^2+n(n+1)x=y^2$$ has always solutions for a given $n>2$. I made the left hand side a square by adding $(n(n+1)/4)^2$ to both sides and get some Pythagorean triple. 
Another way which I could see is after moving $y^2$ to the left hand side and using discriminant $\Delta=(n(n+1))^2+(4y)^2$ and forcing that $\Delta$ must be (at least) a perfect square.
but I like to know if there is a simpler way to determine $x,y$. 
Thanks.
 A: As you say, completing the square:
$$4x^2+n(n+1)x=y^2$$
$$\implies$$
$$\bigg[8x+n(n+1)\bigg]^2-\bigg[4y\bigg]^2=\bigg[n(n+1)\bigg]^2$$
$$\implies$$
A difference of two squares.  Using the identity:
$$N=d_1\cdot d_2=\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2$$
Set
$$d_1 \cdot d_2=\bigg[n(n+1)\bigg]^2$$
The product of consecutive integers is always even, and the square of which is always divisible by $4$:
$$n(n+1)=2\cdot m \qquad \to \qquad \bigg[n(n+1)\bigg]^2=4\cdot m^2$$
But, the only class of numbers that cannot be represented  by a difference of two squares is the $ 2 \pmod 4 $ class, the prime number factorization of which would have a single power of two.  See Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$ 
Since we cannot know beforehand what the power of two of $ \ n(n+1) \ $ will be, other than at least $1$, let
$$n(n+1)=2^r \cdot P \qquad \bigg{|}\qquad r\in \mathbb{N} \ (\min\{r\}=1), \qquad P=\text{an odd product of primes} $$ 
Means that our
$$N=d_1 \cdot d_2=\bigg[n(n+1)\bigg]^2=2^{2r}\cdot P^2$$
then
$$\bigg[8x+2^r \cdot P\bigg]^2-\bigg[4y\bigg]^2=2^{2r}\cdot P^2$$
If we were to starting dividing by $2$, to maintain the difference of squares on the left, we would need to do it an even number of times.  Suppose we do it $2k$ times for some $k\in \mathbb{N}$.  Then we would have
$$\bigg[2^{3-k} \cdot x+2^{r-k} \cdot P\bigg]^2-\bigg[ 2^{2-k} \cdot  y\bigg]^2=2^{2r-2k}\cdot   P^2$$
Note that the exponent on the right is strictly even and cannot reach $1$, which if it did, would result in trying to equate a $2 \pmod 4$ number to a difference of squares, which we know to be an impossible task.
A: For the equation.
$$4x^2+n(n+1)x=y^2$$
It is necessary to factorize.
$$n(n+1)=abc$$
The decision will be.
$$x=\frac{c(b-a)^2}{16}$$
$$y=\frac{c(b-a)(b+a)}{8}$$
