Candies' logic quadratic problem

Now, $x$ and $y$ represents candies, therefore $x,\;y \geq 0$

$ax^2 + ax - 2 = 0\\ x^2 + x - \dfrac{2}{a} = 0$

Now, we proceed to find the value of $x$ :

$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-\frac{2}{a})}}{2(1)}\\ = \dfrac{-1 \pm \sqrt{1 +\frac{8}{a}}}{2}$

Now, for real integers of $x$ , we know that

$1 + \dfrac{8}{a} \geq 0\\ \sqrt{1 +\dfrac{8}{a}} \geq 0\\ -\sqrt{1 +\dfrac{8}{a}} \leq 0\\ \dfrac{-1 - \sqrt{1 +\frac{8}{a}}}{2} < 0$

The equation will have one negative root, therefore, the other root must be positive:

$\dfrac{-1 + \sqrt{1 +\frac{8}{a}}}{2} \geq 0\\ -1 + \sqrt{1 +\dfrac{8}{a}} \geq 0\\ \sqrt{1 +\dfrac{8}{a}} \geq 1\\ 1 +\dfrac{8}{a} \geq 1\\ \dfrac{8}{a} \geq 0\\ a>0$

$a$ is also a positive integer.

Now, we want $x$ to be an integer, therefore

$\sqrt{1 +\dfrac{8}{a}}$ must be an integer, which means

$1 +\dfrac{8}{a}$ is a perfect square (which is also an integer)

For $1 +\dfrac{8}{a}$ to be an integer, $a$ can only take factors of $8$ , therefore

$a=1,\;2,\;4,\;8$

From these 4 values, only one value of $a$ gives a perfect square: $a=1$

Then, $x = \dfrac{-1 + \sqrt{1 +\frac{8}{1}}}{2} =\dfrac{-1 + \sqrt{9}}{2} = \dfrac{3-1}{2} = 1$

$a=1,\;x=1$ . Substitute both of these into the equation $x = ay^2 +ay - 11$ :

$1 = y^2 + y - 11\\ y^2 + y - 12 = 0\\ (y+4)(y-3)=0\\ y=-4,\;3$

Since $y \geq 0,\; y=\boxed{3}$

• $x$ and $y$ can only be integers equal to or larger than 0. As $ax^2+ax-2=0$ , the largest possible value of $x$ is $1$ no matter what value " $a$ " is, such that $1^2+1-2=0$ . Hence, we get $a=1$ too.
• As $x=1$ , $ay^2+ay-11=1$ . Substituting a=1 and x=1 in the equation, we get $y^2+y-12=0$ , where $y=3$ and $-4(rejected)$
• So the answer is 3.