If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots

If the quadratic has rational roots, it can be expressed in the form $$ ax^2+bx+c = (Ax+B)(Cx+D) $$ for integers A, B, C, and D. Expanding and matching, we see that $$ a=AC\qquad b=AD+BC\qquad c=BD $$ For $a$ to be odd, we require $A$ and $C$ to both be odd. Similarly, for $c$ to be odd, we require both $B$ and $D$ to be odd. However, if all of $A$, $B$, $C$, and $D$ are odd, then $AD+BC$ must be even, and thus $b$ must be even.

Thus, to have rational roots, all the coefficients cannot be odd at the same time.

Let the quadratic be $f(x) = ax^2+bx+c$ where $a, b, c \equiv 1 \pmod{2}$. By the Rational Root Theorem, if $\frac{p}{q}$ is a root of the quadratic in its lowest terms, then $p | c$ and $q | a$. Since $a$ and $c$ are odd, then both $p$ and $q$ must be odd. Then, we have $$f(\frac{p}{q}) = a\cdot \frac{p^2}{q^2}+b\cdot \frac{p}{q}+c = \frac{ap^2+bpq+cq^2}{q^2}.$$

However, we have that $a, b, c, p,$ and $q$ are all odd, so then $ap^2+bpq+cq^2$ is also odd, which means we cannot have $f(\frac{p}{q}) = 0$ by contradiction. Therefore, the quadratic $f(x)$ cannot have any rational roots.

(Partly derived from AoPS Algebra 2 textbook)

Take $a=1,b=3,c=2$ to get the rational solutions $-2,-1$. So the statement is false unless $c$ is also required to be odd.

Now consider squares modulo $8$. Any odd number has the form $8n+1$, $8n+3$, $8n+5$, or $8n+7$ (these are abbreviated as $\equiv1,3,5,7\bmod8$). So an odd number squared is

$$1^2=1$$

$$3^2=9=8\cdot1+1\equiv1$$

$$5^2=25=8\cdot3+1\equiv1$$

$$7^2=49=8\cdot6+1\equiv1.$$

And any odd number times $4$ is

$$4\cdot1=4$$

$$4\cdot3=12=8\cdot1+4\equiv4$$

$$4\cdot5=20=8\cdot2+4\equiv4$$

$$4\cdot7=28=8\cdot3+4\equiv4.$$

Therefore, if $a,b,c$ are all odd, then $ac$ is also odd, and

$$b^2-4ac\equiv1-4=-3=8\cdot(-1)+5\equiv5\not\equiv1$$

so $b^2-4ac$ cannot be a square.