Question

The number of integral values of $a$ such that $x^2+ax+a+1=0$ has integral roots is

Answer 1

Given : $x^2+ax+a+1=0$
As the roots are integer, so discriminant should be perfect square,
$D=a^2-4(a+1)\Rightarrow D=a^2-4a+4-8\Rightarrow D=(a-2{)}^2-8$
For integral $a$,
$\Rightarrow D=1\Rightarrow (a-2{)}^2=9\Rightarrow a-2=\pm 3\therefore a=5,-1$

Hence, there are $2$ integral values of $a$.