A problem by Atul Anand Sinha

If we take the consecutive integer numbers $0, 1, 2, 3$ and $0, 1$ , respectively, then the products are equal to zero.

Suppose now that there are integers $x$ and $y$ so that

$x(x+1)(x+2)(x+3)=y(y+1)$

and these products are not equal to zero. Then $y<-1$ or $y>0$ . Note that

$\begin{aligned} \\ x(x+3)&=x^2+3x=(x^2+3x+1)-1 \text{, and} \\(x+1)(x+2)&=x^2+3x+2=(x^2+3x+1)+1 \text{.} \end{aligned}$

Using $(a-b)(a+b)=a^2-b^2$ , we get

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

This means that $y(y+1)$ is one less than a square number, or in other words: $y(y+1)+1=y^2+y+1$ is a square. But we have

$y^2<y^2+y+1<(y+1)^2=y^2+2y+1\text{,}\hspace{17pt}\text{if}y>0$

and

$(y+1)^2=y^2+2y+1<y^2+y+1<y^2\text{,}\hspace{17pt}\text{if}y<-1$ .

This means that unless $y(y+1)=0$ , the number $y(y+1)+1$ is always between two consecutive squares so it cannot be equal to a square. This shows that the original product could only be $\boxed{0}$ .