The equation ax^2+bx+c=0 has a rational solution. Prove that at least a,b, or c is even?

Question

Let a,b, c be integers, and suppose that the eqaution ax^2+bx+c=0 has a rational solution. Prove that at least one of a, b, c must be even.

Answer 1

If ax^2 + bx + c = 0 has a rational solution then

b^2-4ac, the discriminant of the equation, must be a square.

Suppose b^2-4ac= d^ 2 and a,b,c are all odd.

Then b^2 = 1(mod 8) and 4ac= 4(mod 8)

Thus  b^2-4ac= -3 =5(mod 8), so d^2 = 5(mod 8)

But any square is congruent to 0, 1 or 4(mod 8)

so this is impossible.