Question 1027170
Assume otherwise that *[tex \large ax^2 + bx + c] has a rational solution *[tex \large x_0]. Then *[tex \large x_0] is rational if and only if the discriminant *[tex \large b^2 - 4ac] is a perfect square (using the quadratic formula, coupled with the fact that a,b,c are integers).


a,b,c are odd. We will show that *[tex \large b^2 - 4ac] cannot possibly be a perfect square by looking at it modulo 8. All of the odd perfect squares (1^2, 3^2, 5^2, 7^2) leave a remainder of 1 when divided by 8. However *[tex \large b^2 \equiv 1 \pmod{8}] and *[tex \large 4ac \equiv 4 \pmod{8}], so *[tex \large b^2 - 4ac \equiv 1-4 \equiv 5 \pmod{8}], i.e. b^2 - 4ac always leaves a remainder of 5 when divided by 8. Therefore it cannot be a perfect square, and any real solution x cannot be rational.