If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots If the coefficients of a quadratic equation $$ax^2+bx+c=0$$ are all odd numbers, show that the equation will not have rational solutions.
I am also not sure if I should consider $c$ as a coefficient of $x^0$, suppose if I take that $c$ is also odd,
then $$b^2-4ac $$ will be odd. But that $-b$(odd), in the quadratic formula will cancel out the oddness of $\sqrt{b^2-4ac}$, in case if it is a perfect square. If it's not a perfect square then the root is irrational.
If I take $c$ to be even, even then the  same argument runs but we noticed that when we take $c$ odd, we get that when discriminant is perfect square, so that means the question asks for $c$ not to be an coefficient.
Final question: Is this right to take $c$ as one of the coefficients of the equation $ax^2+bx=c=0$?
 A: Let $a=2p+1, b=2q+1, c=2r+1$, where $p,q,r$ are some integers
Then $b^2-4ac=(2q+1)^2-4(2p+1)(2r+1)$
$=4q^2+1+4q-4(4pr+2p+2r+1)$
$=4k-3$
where $k=q^2+q-4pr-2p-2r$,   an even integer
So $b^2-4ac$     is an odd number.So if it is square of some integer , then that integer is odd.
Let $4k-3=(2m+1)^2=4m^2+4m+1$
$\Rightarrow 4(k-m^2-m)=4$
$k-m^2-m=1$
$k=m(m+1)+1$   an odd integer, a contradiction
A: Every odd square is 1, modulo 8.
The square root of an integer $n$ is either an integer (times $i$, if $n<0$) or irrational.
$a, b$ and $c$ are all odd. So, modulo 8, $b^2=1$, $4ac=4$, and $D=b^2-4ac=5$. Thus $D$ is not a square. But $D$ is an integer, so $\sqrt{D}$ is irrational, so the quadratic's roots are not rational.
A: 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.
A: 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)
A: 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.
