Irish Mathematical Olympiad

We would like to find all pairs of integers $(x,y)$ which satisfy $$x(x+1)(x+7)(x+8)=y^2.$$ With $z=x+4$ this equation translates into the equivalent equations $$\begin{array}{rl}& (z-4)(z-3)(z+3)(z+4)=y^2\\ & (z^2-9)(z^2-16)=y^2\\ & z^4-25z^2+12^2=y^2\text{now multiply by 4}\\ & 4z^4-100z^2+24^2=4y^2\\ & (2z^2-25{)}^2-25^2+24^2=4y^2\\ & (2z^2-25{)}^2-(2y{)}^2=49\\ & (2z^2-25-2y)(2z^2-25+2y)=49.\end{array}$$ Because $(x,y)$ is a solution if and only if $(x,-y)$ is so, we may assume $y\ge 0$. This implies $A:=2z^2-25-2y\le B:=2z^2-25+2y$. Because $AB=49$ and $A,B$ are integers, the table below contains all possibilities for them. The values of $y,z,x$ are obtained from $$B-A=4y,2z^2=25+A+2y=25+\frac{A+B}{2}\text{and}x=z-4.$$

A	B	y	z	x
-49	-1	12	0	-4
-7	-7	0	±3	-1, -7
7	7	0	±4	0, -8
1	49	12	±5	1, -9

This gives at most seven possible integers $x$ for which $x(x+1)(x+7)(x+8)$ is a perfect square, namely $x\in \{-9,-4,1\}$ with $y^2=144$ \text{ and } $x\in \{-8,-7,-1,0\}$ with $y^2=0.$ and a quick check reveals that these in fact solve the given equation.