How to prove that $|f(\frac pq)| \geq \frac {1}{q^2}$ for $f(x) = ax^2 + bx + c$ with real roots and coefficients as odd positive integers

The question says:

If $$f(x) = ax^2 + bx + c$$ have real roots and it's coefficients are of positive integers, then

(A) $$f(x) = 0$$ always have real roots

(B) $$\left|f\left(\frac pq\right)\right| \geq\frac {1}{q^2}$$; $$p,q \in I$$

(C) If $$a.c=1$$, then equation must have exactly one root $$\alpha$$ such that $$[\alpha]=-1$$, where $$[.]$$ is greatest integer function

(D) Equation have rational roots

I was able to prove (A) and (C) as right, and (D) as wrong, but can't seem to figure out on how to correctly proceed with (B), which is correct according to the answer key.

I tried using $$A.M. >= G.M.$$ with $$aq^2$$ and $$cq^2$$, along with knowing that $$b^2 > 4ac$$ but didn't seem to get anywhere.

Any hints or suggestions are most welcome for (B).

Thank you!

• Well, $f(p/q)=\frac{M}{q^2}$ for some integer $M$. So . you just need that $M\neq 0,$ which is that $f(p/q)\neq 0.$ So is it possible for $f$ to have a rational root? – Thomas Andrews May 13 '19 at 18:12
• Is $I$ the integers? You said integers previously, just want to make sure you mean the same/ – Thomas Andrews May 13 '19 at 18:14
• @ThomasAndrews but how do we know that $M$ will be an integer? Because as far as I understand, the question is just asking to prove that it $f(p/q)$ is greater than $\frac 1{q^2}$, which may or may not be of the form $M/q^2$. Edit: nevermind, I got it why $M$ will be an integer – Eagle May 13 '19 at 18:17
• @ThomasAndrews yes $I$ is integers – Eagle May 13 '19 at 18:17
• @ThomasAndrews thanks a lot! I understood it now! :) – Eagle May 13 '19 at 18:22

Since $$a,b,c$$ are odd, we have that:

$$f\left(\frac{p}{q}\right)=\frac{ap^2+bpq+cq^2}{q^2}\tag{1}$$

Now, we'll show that, since $$a,b,c$$ are odd integers and $$p,q$$ are integers with $$\gcd(p,q)=1$$, then $$ap^2+bpq+cq^2$$ is always odd, and hence never zero.

If both $$p,q$$ is odd, then $$ap^2+bpq+cq^2=\text{odd }+\text{ odd }+\text{ odd}=\text{odd.}$$

If exactly one of $$p,q$$ is odd, then $$ap^2+bpq+cq^2=\text{even }+\text{ even }+\text{ odd}=\text{odd.}$$

So we have that $$\left|ap^2+bpq+cq^2\right|\geq 1,$$ and hence $$\left|f\left(\frac{p}q\right)\right|=\frac{\left|ap^2+bpq+cq^2\right|}{q^2}\geq\frac{1}{q^2}$$

This also means that $$f$$ has no rational root.

You can prove (C) from (D) if you take another approach to $$D$$.

For example, $$b^2-4ac\equiv 5\pmod{8}$$ when $$a,b,c$$ are odd integers, and thus is never the square of an integer, and thus no root is rational. So from (1) we see that $$|f(p/q)|\geq 1/q^2.$$

\begin{aligned}\left\lvert\frac{ap^2}{q^2}+ \frac{bp}{q} +c\right\rvert &\geq \frac{1}{q^2} \\ \left\lvert\frac{ap^2+bpq+cq^2}{q^2}\right\rvert &\geq \frac{1}{q^2} \\ \frac{|ap^2+bpq+cq^2|}{|q^2|} &\geq \frac{1}{q^2} \end{aligned}

(There's a mistake in my answer—Thomas Andrews has a nice continuation.)

• Can $p$ be negative? – Andrei May 13 '19 at 18:37
• $p,q$ are integers, not necessarily positive. – Thomas Andrews May 13 '19 at 18:39
• So the real question is, can $ap^2+vbpq+cq^2=0$ for $p,q$ integers? – Thomas Andrews May 13 '19 at 18:41