2016/060) Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer (21st Irish math olympiad)

we see that x = 0, x = -1, x = -7 and x = -8 gives the answer zero so a perfect square
let us look for other values
we have
$x(x+1)(x+7)(x+8)$
= $x(x+8)(x+1)(x+7)$
= $(x^2+8x)(x^2+8x+7)$
$= y(y+7)$ where y is $x^2+8x$
for it to be a perfect square we see that $(GCD(y,y+7) = GCD(y,7)$
y cannot be a multiple of if 7 beacuse then y and y + 7 are consecutive multiples of 7 and as y is not zero product cannot
be a perfect square.
so y and y + 7 are coprimes and hence perfect square or -ve of perfect square
let $y = n^2$ and $y+7 = m^2$
giving $n^2+7=m^2$
or $m^2-n^2 = 7$
or $(m+n)(n-n) = 7 * 1$ hence $m+n = 7, m-n= 1$ or $m= 4,n= 3$
so $y = 16$
hence $x^2+8x-9=0$ giving $x = 1,=9$
taking -ve values we have $y= - 16 , y + 7 = - 9$
or $x^2+8x+ 16= 0 => x = - 4$
so we have x is one of -9,-8,-7, -1,0,1$