Polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ & $c$ are positive integers) Let the polynomial $f(x) = ax^2 – bx + c $ (where $a$, $b$ &amp; $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 &lt; p &lt; q &lt; 1$, then find the minimum possible value of $a$.
The vertex is $-\frac{-b}{2a}=\frac{b}{2a}&gt;0$ and lies between $0$ &amp; $1$.
$f(0)&gt;0$ and also $f(1)&gt;0$, hence $c&gt;0$ and $a-b+c&gt;0$, also $b^2-4ac&gt;0$. 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 $f(x)=5x^2-5x+1$
You have $ax^2-bx+c = a(x-p)(x-q)=ax^2-a(p+q)x+apq$. So $a(p+q)=b\geq 1$ and $apq=c\geq 1$. This implies $a\geq 1/(p+q)$ and $a\geq 1/pq$. But your constraints on $pq$ force $1/pq&gt;1/(p+q)$. So really the bound is:
$$
a\geq 1/pq.
$$
Without more information there is no improvement.
For example let $p=1/3$ and $q=1/2$. Then You can take $a=6$, $b=5$, and $c=1$.