2016/056) If $n^4+an^3+bn^2-8n+1$ is a perfect square for all integer values of n. find a,b

$n^4+an^3+bn^2-8n+1$ is a perfect square for all integer values of n so this is square of a polynomial.
hence
$n^4+an^3+bn^2-8n+1 = (n ^2 + mn + c)^2$
comparing the constant term we have $1=c^2$ or $c = \pm 1$
so we need to consider 2 cases
c = 1 and c= -1
case 1
c = -1
this give $(n^2 + mn + 1)^2 = n^4 + mn^3 + n^2(m^2+2) + mn +1 = n^4+an^3+bn^2-8n+1$
comparing coeffcients on both sides $m = a, b = m^2 + 2, - m = -8$ giving $a= 8, b = 66$
case 2
c = -1
this give $(n^2 + mn - 1)^2 = n^4 + mn^3 + n^2(m^2-2) - mn +1 = n^4+an^3+bn^2-8n+1$
comparing coefficients on both sides $m = a, b = m^2 - 2, - m = -8$ giving $a= -8, b = 62$
so we have 2 sets $a=8, b= 62$ and $a= -8,b=66$