Conclude and Prove

Yes, but then you would still have to deal with the case $n = 1$ , which is why I combined the cases in saying that for $n \ge 1$ we have $q(1 - n^{2}) \le 0$ .

For there to be rational solutions we would require that the discriminant be a perfect square. So suppose that $b^{2} - 4ac = n^{2}$ for some positive integer $n.$

Note first that, with $a,b,c$ all odd, we would have $b^{2} - 4ac$ being odd, and thus $n$ would be odd as well. Now rewrite the equation as

$b^{2} - n^{2} = 4ac \Longrightarrow (b - n)(b + n) = 4ac.$

Now since $a,c$ are both odd, $4ac$ has exactly two factors of $2.$ For the left-hand side, since $b,n$ are both odd, both $(b - n)$ and $(b + n)$ are even and differ by $2n.$ Now since $n$ is odd, we know that $2n \equiv 2 \pmod{4}.$ Thus one or the other of $(b - n)$ or $(b + n)$ must be divisible by $4,$ which implies that $(b - n)(b + n)$ has at least three factors of $2.$ Since $4ac$ has exactly two factors of $2,$ we can conclude by contradiction that there exists no positive integer $n$ such that $b^{2} - 4ac = n^{2}$ where $a,b,c$ are all odd, and thus $ax^{2} + bx + c = 0$ has no rational roots when $a,b,c$ are all odd, positive integers.

In this case, the discriminant is $23^{2} - 4*3*23 = 253$ , which is greater than zero but is not a perfect square. The roots of the equation are

$x = \dfrac{23 \pm \sqrt{253}}{6}$ ,

which are not rational.

My pleasure, Manish. Thanks for the compliment. :)

The terms rational and irrational only apply to real numbers. A complex number $a + bi$ can have rational and irrational parts $a,b$ but we wouldn't generally refer to the complex number itself as being either rational or irrational. I've shown that any real solution will not be rational, and from my previous comment any complex solution will by definition neither be rational nor irrational, but just simply complex.

Now we could have a complex solution that has rational parts, however. For example, with $p = 5, q = 2$ we have the equation $5x^{2} - 2x + 2 = 0$ , which has solutions

$x = \dfrac{2 \pm \sqrt{4 - 40}}{10} = \dfrac{2 \pm 6i}{10} = \dfrac{1}{5} \pm \dfrac{3}{5}i$ .

I was thinking the same thing.

The polynomial $x^{2} - 3x + 2$ is not in the desired form $px^{2} - qx + q$ for primes $p,q.$ Also, regarding your first sentence, while it is true that $AB = q$ this does not mean that $A$ or $B$ has to be $1,$ since we could be looking for rational roots, (and not necessarily integer roots).