Polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers) Let the polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 < p < q < 1$, then find the minimum possible value of $a$.
The vertex is $-\frac{-b}{2a}=\frac{b}{2a}>0$ and lies between $0$ & $1$.
$f(0)>0$ and also $f(1)>0$, hence $c>0$ and $a-b+c>0$, also $b^2-4ac>0$. Even after proceeding up to these steps I am not able to find the minimum value of $a$.

Let the quadratic be $a(x-u)^2-v$, where $u=(p+q)/2$ and $-v=f(u)\lt0$.
Then $f(0)=au^2-v\ge1$ so $au^2\gt1$ and likewise $a(1-u)^2\gt1$.  Either $u\le\frac12$ or $1-u\le\frac12$ so $\frac a4\gt1$, and $a\ge5$.
An example is $f(x)=5x^2-5x+1$
