Integer solutions to the product of four consecutive integers

Question

$ \bullet \textbf{Question} $

The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.

$ \bullet \textbf{Rephrasing} $

I decided to name the other integer as $ y $ so that this equation is a plot on a graph,

$$ y(y + 1) = x(x + 1)(x + 2)(x + 3) \tag{1} $$

$ \bullet \textbf{Attempt} $

To solve for $ y $ I did these simple steps,

$$ y^2 + y = x(x + 1)(x + 2)(x + 3) \tag{2} $$

$$ y^2 + y - x(x + 1)(x + 2)(x + 3) = 0 \tag{3} $$

$$ y = \frac{-1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}}}{2} \tag{4} $$

I then figured that when,

$$ -1 \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 0\pmod{2} \tag{5} $$

then $ y $ will be a integer, so this is just solving for all $ x $, so $ y $ will be a integer but not for only integer $ x $'s.

This then means that

$$ \pm{\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{6} $$

$$ {\sqrt{1 + 4x(x + 1)(x + 2)(x + 3)}} \equiv 1\pmod{2} \tag{7} $$

This is the end of my knowledge, if you try to square it will include the integer solution for $ y $ but also include non-integer solutions for $ y $.

Am I going in the correct direction, or is this a dead end and I need to apply a new approach?

Answer 1

$x(x + 1)(x + 2)(x + 3)=(x^2+3x)(x^2+3x+2)$.

Let $N=x^2+3x+1$, then $N^2-1=m(m+1)$.

If $N=0$

There are no solutions since $m(m+1)$ is even.

If $|N|=1$

Then $m=0$ or $-1$.

If $|N|\ge 2$

Then $N^2-1$ is greater than $|N|(|N|-1)$ but less than $|N|(|N|+1)$. There are no solutions.

The only solutions are when $x(x + 1)(x + 2)(x + 3)=0$ i.e. $x\in\{-3,-2,-1,0\}$.

Answer 2

We have $$y^2+y=(x^2+3x)(x^2+3x+2)$$ or$$y^2+y+1=(x^2+3x+1)^2$$ or $$(2y+1)^2+3=(2x^2+6x+2)^2$$ or $$(2x^2+6x-2y+1)(2x^2+6x+2y+3)=3.$$ Now, solve a number of systems.