<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] The equation of the parabola is:
$$a \left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$$
$$=a \left(x^2-\frac{1}{2}x+\frac{1}{16} \right)- \frac{9}{8}$$
$$=ax^2-\frac{a}{2}x+ \left(\frac{a}{16}-\frac{9}{8} \right)$$
Then we find that: $$a+b+c = \frac{9}{16}(a-2).$$ 
When $a=0$, $a+b+c = -\frac{9}{8}$. Since the gradient of the line is positive, if we try $a+b+c=-1$, we will get a positive value of $a$.
Therefore we have that: $-1 = \frac{9}{16}(a-2) \Rightarrow a = \frac{2}{9}$.
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