TITLE: If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots
QUESTION [9 upvotes]: 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 $ax^2+bx=c=0$?

REPLY [5 votes]: 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.