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

Question

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 $N(6)$?

Answer 1

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 $x|6$.

Answer 2

First of all thanks André Nicolas for the hint.

Using $ Vieta's \space formula$, we get

Product of roots $=x_1x_2=b=6=3\times 2=1\times6$

Sum of roots $=x_1+x_2=-a$

As $x_1, x_2$ are integers, $ x_1=-6, x_2=-1$ or $x_1=-3,x_2=-2$ (we may also interchange the values of $x_1,x_2$ but that will not effect the values of $a$. We also cannot use the positive values as they will make $a$ negative).

Hence, $a$ has two possible values $7$ and $5$.