Let $m,\:n\:∈\:Z$. Show that If $\:x^2\:+\:mx+n\:=\:0$ has a rational root, then m, n cannot be both odd.

Question

Here's my attempt, would this be considered a valid proof?

Proof by contradiction: Assume m and n are both odd integers and the equation $\:x^2\:+\:mx+n\:=\:0$ has a rational root.

• m = $2t+1$ (for some $t\in \mathbb{Z}$)
• n = $2s+1$ (for some $s\in \mathbb{Z}$)

Rewrite equation: $x^2+\left(2t+1\right)x+\left(2s+1\right)=0$

Solve for $t$: $t=\frac{-x^2-2s-1-x}{2x};\quad \:x\ne \:0$

Solve for $s$: $s=\frac{-x^2-2xt-x-1}{2}$

From this it is clear that $t \notin \mathbb{Z} $ and $s \notin \mathbb{Z}$ and so $m$ and $n$ are also $\notin \mathbb{Z}$.

This shows that for the equation to have a rational root, if m and n are odd, they both cannot be integers.

And so, if $m\in \mathbb{Z}$ and $n\in \mathbb{Z}$, they both CANNOT be odd, as required.

$∎$

Answer 1

If $x^2+mx+n$ has a rational root, then both roots are rational. So multiplying by the least common multiple of the denominators we have an equation with integer coefficients where even the roots maintain the same relationship with the coefficients. Suppose the roots of this new equation are the integers $-a$ and $-b$ such that $m=a+b$ and $n=ab$.

Suppose both $m$ and $n$ are odd. Hence $ab=n=odd$, which implies that $a$ and $b$ are odd, that is $a=2\tilde{a}+1$ and $b=2\tilde{b}+1$ for some $\tilde{a},\tilde{b}\in\mathbb Z$. Thus $$\underbrace{m}_{odd}=a+b=\underbrace{2(\tilde{a}+\tilde{b}+1)}_{even}.$$

Hence, $m$ and $n$ cannot be odd at the same time.

Answer 2

"From this it is clear that ..." - even if it was true up to this point, $t$ and $s$ could be fractions with denominator $2$ and $m=2t+1$, $n=2s+1$ would still be integers.

Instead, let's assume $m,n$ are odd and consider the quadratic formula,

$$ x=\frac{-m\pm \sqrt{m^2-4n}}{2}$$

If one root is rational then $m^2-4n$ is a square number, $k^2$, and in fact both roots are rational. We know $m$ is odd and $4n$ is even, so $m^2-4n=k^2$ is odd and thus $k$ is odd (an even number squared is even, but an odd number squared is odd).

Consider the difference between two arbitrary odd square numbers, $(2a+1)^2$ and $(2b+1)^2$. Is this difference even or odd? What contradiction does this give us?

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EDIT:

The difference between two odd square numbers, $(2a+1)^2$ and $(2b+1)^2$, is:

$$ (2b+1)^2-(2a+1)^2\\ =4b^2+4b+1-(4a^2+4a+1)\\ = 4(b^2+b-(a^2+a))$$

$b^2$ is the same parity as $b$ so $b^2+b$ is even. Identically, $a^2+a$ is even and $b^2+b-(a^2+a)$, the difference of two even numbers, is even.

Thus, the difference between any two odd square numbers, $4(b^2+b-(a^2+a))$, is always a multiple of $8$ (it is $4$ times some even number).

Back to our values $m,n$, we know that $m^2-4n=k^2$, where $m,k$ are odd integers - that is, the difference between two odd square numbers is $4n$ (by rearranging, $m^2-k^2=4n$). We have just proved that $m^2-k^2$ must be a multiple of $8$ i.e. that $4n$ is a multiple of $8$, which means that $n$ is even.

But we assumed that $n$ is odd, so this is a contradiction. Hence, $m$ and $n$ cannot both be odd.

Answer 3

As noted in the comments and other answer, the proof is not valid. Here is an partial alternative that uses the rational roots theorem:

Suppose, to the contrary, that both $n$ and $m$ are odd. Let $a_1, a_2, \dots, a_k$ be all of the factors of $n$. Note that $a_i$ must be odd for all $i=1,2,\dots, k$. Then, by the rational root theorem, $a_1, a_2, \dots, a_k$ are all of the possible rational roots of our quadratic, so it turns out that if a root is rational, then it is an integer. Since one of the roots are indeed rational, there exists $j\in\{1,2,\dots,k\}$ such that $a_j^2+ma_j+n=0$. Solving for $n$ gives us $n=-a_j^2-ma_j$. Can you find why this results in a contradiction?