For a natural number $b$, $N(b)=$ number of natural numbers $ \ a \ $ such that the equation $x^2+ax+b=0$ has integral roots. For a natural number $b$, $N(b)= $ number of natural numbers $a$ such that the equation $x^2+ax+b=0$ has integral roots. What is the lowest possible value of $N(6)$?

Rote approach. The solutions to $x^2+ax+6=0$ are $x_1=(1/2)(\sqrt{a^2-24}-a), x_2=(1/2)(-\sqrt{a^2-24}-a)$. First, $\sqrt{a^2-24}$ is rational iff it is integral, so we must have $a=\sqrt{c^2+24}$ for some integral ${c}$. Notice that if $c>12$ this expression can't be an integer, as the difference between consecutive square numbers is larger than 24 at that point, so you need only check $c = -4, -3, ..., 12$.
For a shortcut, note that if $x^2+ax+6=0$ then $x|6$.
