2016/064)Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003$(Irish 2003 paper 2)

Add 1 to both sides to get
$(y+1)^2 = x^4+20x^3+104x^2+40x+2004 = x^4+20x^3+ 104x^2+40x+ 4 + 2000 $
$= (x^2+10x+2)^2 +2000$
or $(y+1)^2 - (x^2+10x+2)^2 = 2000$
for the above to have solution we need to have $(y+1)$ and $(x^2+10x+2)$ both should be positive and as $(x^2+10x+2) = (x+5)^2-23$
so the lower number need to be 23 less than a perfect square.
so let us find (t,z) which are $(\pm501,\pm499),(\pm252,\pm248),(\pm129,\pm121),(\pm105,\pm95),(\pm60,\pm40),(\pm45,\pm5)$ out of which
only z = 121 which 23 less than is a perfect square
so we get
$y+1= \pm 129, (x+5) = \pm 12$ giving 4 solutions ($y=-130, x = - 17$), ($y= -130,x= 7$), ($y=128,x=7$),($y= 128, x= -17$)