An algebra problem by Yash Choudhary

Let $p, q$ be the roots.

Consider $c=apq<a$ .

Then $b=a (p+q) <apq+a=c+a$ as $a (1-p)(1-q) > 0$ .

Also $b > 2\sqrt {pq} a=2\sqrt {ac}$ . This is by AM-GM inequality.

So, now $b$ is bounded between two numbers. One of them, $c+a$ is an integer. This means that the other, $2\sqrt {ac} <c+a-1$ or equivalently $\sqrt {a}-\sqrt {c}> 1$ . This is by bringing $2\sqrt {ac}$ across and square rooting.

(Otherwise $b$ would be between 2 consecutive integers, impossible)

As $c \geq 1$ , this immediately implies $a>4$

Thus the minimum is at least 5. But the quadratic $5x^{2}-5x+1$ satisfies the condition. Thus 5 is the answer.