Prove the following proposition: If a, b and c are integers such that a is odd,

Question

Prove the following proposition:
If a, b and c are integers such that a is odd, b and c are even, but c is not divisible by 4, then the equation ax^2 + bx + c = 0 has no rational solutions.

Assume that every nonzero rational number may be expressed as the quotient of two integers that have no common factors
greater than 1.

Explanation / Answer

Suppose x = r/s is a rational solution to this equation. As noted, we may assume that (r,s) =1. Plugging this into the equation and clearing denominators we have ar^2 + brs + cs^2 = 0. Since b, c are even it follows that ar^2 is even. Since a is odd, we must have that r is even => ar^2 is divisible by 4 but we know that c is not divisible by 4, so reading this equation modulo 4, note that br is divisible by 4, so cs^2 is also divisible by 4 => 4|s^2 which implies that 2|s. But this contradicts the assumption that (r,s) = 1 since we have already concluded that 2|r.

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