Odd quadratic

This is a quite interesting result. Here is a collection of few proofs. But let me type out my approach :

Proof :

Since $a,b,c$ are odd integers, let $a=2x+1 , b=2y+1$ and $c=2z+1$ for $x,y,z \in \mathbb{Z}$ .

Now, Discriminant of the expression $= b^{2} - 4ac = (2y+1)^{2} - 4(2x+1)(2z+1) = 4y^{2} + 4y -16xz -8z -3$

$= 8 \left[ \dfrac{y(y+1)}{2}-2xz-x-z \right] -3 = 8k-3$ for some integer $k$ .

But all odd perfect squares are of the form $8k+1$ .

Hence, Discriminant of the expression $ax^{2} + bx + c$ is not a perfect square, i.e. it doesn't have rational roots. Q.E.D

Suppose x = p/q is a rational root to the above quadratic ax^2 + bx + c = 0. This would yield:

a p^2 + bpq + c q^2 = 0 (i).

Assuming p and q are coprime integers (i.e. both are not even integers):

CASE 1: p & q both odd => (i) is the sum of three odd integers which is strictly an odd integer (contradiction).

CASE 2: p is even(odd) & q is odd(even) => (i) is the sum of two even integers and one odd integer which is strictly an odd integer (contradiction).

Hence, ax^2 + bx + c = 0 can't ever have rational roots for odd a, b, and c. QED