<s>[INST] Parabola equation in form of quadratic $ax^2+bx+c=y$ where $a+b+c$ is an integer Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach
$(x-\frac{1}{4})^2=4a'(y+\frac{9}{8})$
$\frac{x^2-\frac{x}{2}+\frac{1}{16}}{4a'}-\frac{9}{8}=y$
$\frac{1}{4a'}-\frac{1}{4a'}+\frac{1}{64a'}-\frac{9}{8}=Z$, where $Z$ is an integer
$\frac{9}{64a'}-\frac{9}{8}=Z$, from here I am confused
 [/INST] We know that the vertex x coordinates is $\frac{-b}{2a} = \frac{1}{4}$ solve for $b = \frac{-a}{2}$ now the point $\frac{1}{4} , \frac{-9}{8}$ satisfies the parabola so 
$\frac{a}{16} - \frac{a}{8} + c = \frac{-9}{8}$ 
$a-2a + 16c = -18 $ 
$c = \frac{-18 +a}{16}$ finally 
$a+b+c = a + \frac{-a}{2} +\frac{-18+a}{16} \in \mathbb{Z}$  find the solution and pick the smallest one 
we will have $16 \mid 9a-18 $ 
</s>