Question 1017072
You probably mean *rational* solution (since "fraction" is somewhat ambiguous).


The roots of the equation are *[tex \large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}]. It suffices to prove that *[tex \large b^2 - 4ac] cannot be a perfect square, otherwise the solutions would be rational.


If *[tex \large b^2 - 4ac] is a perfect square, then *[tex \large b^2 - 4ac = n^2 \leftrightarrow b^2 - n^2 = 4ac] where a, b, c are odd integers and n is an integer. Note that n is odd, since b^2 is odd and 4ac is even.


Here, we use a little modular arithmetic. The right hand side must leave a remainder of 4 when divided by 8, since a and c are odd. However, all of the odd squares leave a remainder of 1 when divided by 8, meaning that their difference is a multiple of 8. Since the remainders upon division by 8 are not equal, the expressions *[tex \large b^2 - n^2] and *[tex \large 4ac] cannot possibly be equal, and there is no solution in odd integers a,b,c. Therefore *[tex \large b^2 - 4ac] cannot be a perfect square, no rational solution.


In modular arithmetic terms, we say that *[tex \large 4ac \equiv 4 \pmod{8}] and *[tex \large b^2 - n^2 \equiv 0 \pmod{8}].