Proof that ax^2 + bx + c has No Rational Zeroes if a,b, and c are Odd

[IHateFactorial]

Well, let's look at how this works.

Quadratic equations can have either 1, 2, or no zeroes. If it has no real zeroes, the zeroes it DOES have are complex, so that's obviously not it.

Let's imagine \(\displaystyle ax^2 + bx + c = 0\) has one zero, call it \(\displaystyle \alpha\) (Cuz it looks pretty).

Then that means \(\displaystyle ax^2 + bx + c = (x+\alpha)(x+\alpha) = x^2 + 2\alpha x + \alpha^2\)

Seeing that, we see that \(\displaystyle b = 2\alpha\), which isn't odd an odd number.

The above example is imagining \(\displaystyle \alpha < 0\); if \(\displaystyle \alpha > 0\), it'd be the same thing: \(\displaystyle b = -2\alpha\)

Now, let's take it as if it had two zeroes: \(\displaystyle \alpha\) and \(\displaystyle \beta\).

\(\displaystyle ax^2 + bx + c = (x+\alpha)(x+\beta) = x^2 + x\beta + x\alpha + \alpha\beta = x^2 + x(\alpha+\beta) + \alpha\beta\)

So, let's imagine now that all coefficients are odd.

a = 1, because the coefficient of x squared is 1.
b = \(\displaystyle \alpha + \beta\), so one of those most be even not both.
c = \(\displaystyle \alpha \beta\), this leads to a contradiction with b, because if either alpha or beta is even, then b is even.

Is this right? If I show this to my teacher, will she accept it?

 

[Deveno]

Well, let's look at how this works.

Quadratic equations can have either 1, 2, or no zeroes. If it has no real zeroes, the zeroes it DOES have are complex, so that's obviously not it.

Let's imagine \(\displaystyle ax^2 + bx + c = 0\) has one zero, call it \(\displaystyle \alpha\) (Cuz it looks pretty).

Then that means \(\displaystyle ax^2 + bx + c = (x+\alpha)(x+\alpha) = x^2 + 2\alpha x + \alpha^2\)

This is untrue. For example, $x^2 - 2x + 1$ and $4x - 8x + 4$ both have the single root 1, but the coefficient of the $x^2$ term in the second polynomial isn't $a = 1$.

Furthermore, if $\alpha$ is our single root, all we know is that $(x - \alpha)^2$ is a factor (not $(x + \alpha)^2$).

The problem states that the roots under consideration are RATIONAL, so there is reason to suppose $\alpha$ is an integer, for example, the polynomial $4x^2 - 4x + 1$ has integer coefficients, but its only root is $\frac{1}{2}$.

Seeing that, we see that \(\displaystyle b = 2\alpha\), which isn't odd an odd number.

It could be if $\alpha = \dfrac{2k+1}{2}$ for an integer $k$.

The above example is imagining \(\displaystyle \alpha < 0\); if \(\displaystyle \alpha > 0\), it'd be the same thing: \(\displaystyle b = -2\alpha\)

Now, let's take it as if it had two zeroes: \(\displaystyle \alpha\) and \(\displaystyle \beta\).

\(\displaystyle ax^2 + bx + c = (x+\alpha)(x+\beta) = x^2 + x\beta + x\alpha + \alpha\beta = x^2 + x(\alpha+\beta) + \alpha\beta\)

So, let's imagine now that all coefficients are odd.

a = 1, because the coefficient of x squared is 1.
b = \(\displaystyle \alpha + \beta\), so one of those most be even not both.
c = \(\displaystyle \alpha \beta\), this leads to a contradiction with b, because if either alpha or beta is even, then b is even.

Is this right? If I show this to my teacher, will she accept it?

I seriously doubt it.

I think you're lost in useless territory here, what you want to show is that:

$ax^2 + bx + c$ has a rational root if and only if $b^2 - 4ac$ is a perfect square of a rational number. Since $a,b,c$ are all integers, this means $b^2 - 4ac$ is the square of an integer.

If $a,b,c$ are all ODD, then $b^2 - 4ac$ must likewise be an odd integer, and thus if it is a square, is the square of an odd integer.

See if you can prove $b^2-1 = (b+1)(b-1)$ is an odd multiple of $4$. Why is this a contradiction?