Integral roots

If $x$ and $y$ are the roots of the quadratic.Using Vieta's formula, $x+y=-a$ , $xy=3a\implies 3x+3y=-xy$

We can factor the last equation as $xy+3x+3y=0$ $(x+3)(y+3)=9$ It's not to hard to find that the solutions in x and y to this are $(0,0),~(-2, 6),~(-4, -12),~(-6,-6)$

This yields $a=16,~12,~0,~-4$ since we only want the positive our solutions are $a=\boxed{12,~16}$

So sum of all such $a=28$ .

The roots of $x^2+ax+3x=0$ are $x = \dfrac {-a \pm \sqrt{a^2-12a}}2$ . For the root $x$ to be an integer, $a$ and $\sqrt{a^2-12a}$ divisible by 2 and $a^2-12a$ , a perfect square . Let $a=2b$ , where $b$ is a positive integer. Then we have:

$\begin{aligned} \sqrt{a^2-12a} & = 2\color{#3D99F6}c & \small \color{#3D99F6} \text{where }c \text{ is a positive integer.} \\ \sqrt{4b^2-24b} & = 2c \\ 4b^2-24b & = 4c^2 \\ b^2-6b & = c^2 \\ b^2-6b - c^2 & = 0 \\ \implies b & = \frac {6 \pm \sqrt{36+4c^2}}2 \\ b & = 3 \pm \sqrt{9+c^2} \end{aligned}$

For $b$ to be a positive integer, $9+c^2$ must be a perfect square. A trivial solution is $c=0$ , then $b = 0, 6$ $\implies a = 12 > 0$ $\implies x = -6$ , an integer.

Another trivial solution is the Pythagorean triple $3^2 + 4^2 = 5^2$ , therefore, $c=4$ , then $b = -2, 8$ $\implies a = 16 >0$ $\implies x = -4, -16$ , integers.

And there is no other Pythagorean triple possible. Therefore, there are only two solutions for $a$ and their sum is $=12+16=\boxed{28}$ .

Relevant wiki: Quadratic Discriminant - Problem Solving

Since the quadratic equation has integer roots, then its quadratic discriminant must be a perfect square . That is, $a^2 - 4(3a) = a^2 - 12a$ can be expressed as $b^2$ for some non-negative integer $b$ .

By completing the square , we see that

$a^2 - 12a = b^2 \quad \Leftrightarrow \quad (a-6)^2 - 6^2 =b^2 \quad \Leftrightarrow \quad (a+b-6)(a-b-6) = 36 .$

Since $a+b-6$ and $a-b-6$ differ by $2b$ , which is an even number, then both of these numbers must be even as well such that their product (36) is also an even number.

Because $b$ is a non-negative integer, we have $a+b - 6 \geq a -b - 6$ .

Let's write out all the possible ways 36 can be expressed as product of 2 even integers:
$36 = -2\times-18 = -6 \times -6 = 6\times 6 = 2\times 18 .$

So we have 4 cases:
Case 1: $a +b- 6 = -2, a - b - 6 = -18$ . Solving them simultaneously gives $a = -4, b = 8$ .
Case 2: $a +b- 6 = -6, a - b - 6 = -6$ . Solving them simultaneously gives $a = b=0$ .
Case 3: $a +b- 6 = 6, a - b - 6 = 6$ . Solving them simultaneously gives $a =12, b=0$ .
Case 4: $a +b- 6 = 18, a - b - 6 = 2$ . Solving them simultaneously gives $a =16, b=8$ .

Notice that both in Case 1 and Case 2, we have $a\leq 0$ , which are neglected because we are given that $a > 0$ .

Thus, we only have 2 cases left. Adding all the possible values of $a$ gives the answer $12 + 16 = \boxed{28}$ .