# Condition For Integer Roots

Find the number of values of a such that the eauation$$x^2+ax+6a$$ has integer roots. $$x=\frac{-a\pm\sqrt{a^2-24a}}2$$ let$$a^2-24a=m^2$$where m$\in\Bbb{Z}$ $$(a-12)^2=m^2+12^2$$ $$a=12\pm\sqrt{m^2+144}$$ now by putting different values of $m$ $(0,5,35)$ we get the number of values of $a$ as as 6. Is there any other method where I don't have to check the values of $m$ ?

we get the number of values of $a$ as as 6.

I don't think this is correct. For example, $m=16$ works.

Is there any other method where I don't have to check the values of $m$ ?

We have $$a=\frac{-x^2}{x+6}=-x+6-\frac{36}{x+6}$$ Since $a$ has to be an integer, $$\frac{36}{x+6}$$ has to be an integer. So, $x+6$ has to be a divisor of $36$.

Note here that the discriminant is $a^2-24a$, so considering the cases where $a=0$ or $a=24$ gives : for $a=0$, $x=0$, and for $a=24$, $x=-12$.

Now since the number of the divisors of $36$ is $18$, the number of $a$ we want is $$1+1+\frac{18-2}{2}=\color{red}{10}$$

• @NoChance : In this answer, $10$ represents the number of values of $a$ such that $x^2+ax+6a=0$ has integer roots. Commented Oct 11, 2023 at 14:20
• @Splended, Thanks for the quick reply. Commented Oct 11, 2023 at 23:19

You can find all the possible $m$ by determining all divisors $d$ of $144$ (also the negative divisors!) and solve the linear system

$$a-12-m=d$$

$$a-12+m=\frac{144}{d}$$

Whenever $a$ and $m$ are both integers, you have found a solution.

To answer the given question : You have to do brute force, but in this case, you find the solutions fast.

• It turns out that exactly the even $d$, such that $\frac{144}{d}$ is even, do the job. Commented Aug 31, 2016 at 17:22