Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.

Hint $\ $ By the Rational Root Test, any rational root is integral, hence it follows by
Theorem Parity Root Test $\ $ A polynomial $\rm\:f(x)\:$ with integer coefficients 
has no integer roots if its constant coefficient and coefficient sum are both odd. 
Proof $\ $ The test  verifies that $\rm\  f(0) \equiv  1\equiv f(1)\ \ (mod\ 2)\:,\ $ i.e. 
that $\rm\:f(x)\:$ has no roots modulo $2$, hence it has no integer roots. $\ $ QED
This test extends to many other rings which have a "sense of parity", i.e. an image $\cong \Bbb Z/2,\:$ for example, various algebraic number rings such as the Gaussian integers.
