Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd

Question

If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?

I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and rational numbers are of the form $\frac pq$.

Answer 1

You should think about factoring the polynomial rather than finding its roots. If $ax^2 + bx + c$ has a rational root, then its other root must also be rational, and then it factors in some way like this: $$ax^2 + bx + c = (Ax + B)(Cx + D) \quad A,B,C,D \in \mathbb{Z}$$ Then $a = AC$ and $c = BD$, so if both are odd, then all of $A,B,C,D$ are odd. But then we also have $b = AD + BC$, which is the sum of odd integers, and therefore is even.

Answer 2

Consider a quadratic equation of the form $a\cdot x^2 + b\cdot x + c = 0$. The only way, it can have rational roots IFF there exist two integers $\alpha$ and $\beta$ such that

$$\alpha \cdot \beta = a\cdot c\tag1$$ $$\alpha + \beta = b\tag2$$ $$ Explanation\left\{ \begin{align} if\,\alpha\cdot \beta &= a\cdot c,\\ \frac{\alpha}{a} &= \frac{c}{\beta}\\ a\cdot x^2 + b\cdot x + c& = a\cdot x^2 + (\alpha + \beta)\cdot x + c\\ & = a\cdot x^2 + \alpha\cdot x + \beta\cdot x + c\\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{c}{\beta}))\\ & =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{\alpha}{a}))\\ & =(x+\frac{\alpha}{a})\cdot (a\cdot x + \beta)\\ &\text {As a Quadratic equation has only two roots,}\\ &\text {there would be no other way to factorize the equation} \end{align}\right. $$ $$\text{Reason }\alpha,\beta\in\mathbb{Z}\begin{cases} \text{Given a Ring R, with two operations }\left\{⋅,+\right\},\text{ on }\mathbb{Q}\\ \text{and if } \alpha,\beta \in \mathbb{Q},\alpha\cdot \beta \in \mathbb{Z},\alpha+\beta\in\mathbb{Z}\\ \Rightarrow\alpha,\beta \in \mathbb{Z}\\ \text{ where } \mathbb{Z}\subset\mathbb{Q} \end{cases} $$ If $(a,b,c)$ are odd then $\alpha \cdot \beta$ is odd and $\alpha + \beta$ is odd, but you cannot have two integers whose product and sum are odd.

So by contradiction we prove that the equation cannot have rational roots

Answer 3

Suppose that $a,b,c$ are odd. $ax^2+bx+c=0$ has rational roots iff the discriminant is the square of an integer. That is, there is an integer $d$ so that $d^2=b^2-4ac$. Since $a,b,c$ are odd, $d$ must also be odd.

Note that the right hand side of $$ (b-d)(b+d)=4ac\tag{1} $$ has exactly two factors of $2$. However, since $b$ and $d$ are both odd, $2d\equiv2\pmod{4}$ and so one of $b-d$ and $b+d$ is $0\bmod{4}$ and the other is $2\bmod{4}$. Thus, the left hand side of $(1)$ has at least three factors of $2$. Contradiction.

Answer 4

By Gauss's lemma It suffices to consider the domain of $x$ as the integers. If $x$ is even or odd then $ax^{2}+bx+c=x(ax+b)+c$ is odd hence not equal to $0$.

Answer 5

Hint $\ $ By the Rational Root Test, any rational root is integral, hence it follows by

Theorem Parity Root Test $\ $ A polynomial $\rm\:f(x)\:$ with integer coefficients has no integer roots if its constant coefficient and coefficient sum are both odd.

Proof $\ $ The test verifies that $\rm\ f(0) \equiv 1\equiv f(1)\ \ (mod\ 2)\:,\ $ i.e. that $\rm\:f(x)\:$ has no roots modulo $2$, hence it has no integer roots. $\ $ QED

This test extends to many other rings which have a "sense of parity", i.e. an image $\cong \Bbb Z/2,\:$ for example, various algebraic number rings such as the Gaussian integers.

Answer 6

If $\frac{p}{q}$ is a root then $a\frac{p^2}{q^2}+b\frac{p}{q}+c=0\Rightarrow ap^2+bpq+cq^2=0$. Now we may assume that $p,q$ are not both even; $a,b,c$ are odd, whence contradiction ($p(ap+bq)=-cq^2$ with $p$ even $c,q$ odd or $q(bp+cq)=-ap^2$ with $q$ even $a,p$ odd or ...).

Answer 7

You can actually prove this in quite an elementary way without even knowing anything about the roots of the quadratic equation. Suppose, on the contrary, that we have rational root $\dfrac {p}{q}$. Then your equation is equivalent to $ap^2+bpq+cq^2=0$. You have that $a,b,c$ are all odd. I will denote odd as an $O$ and even as an $E$.

This breaks into four cases.

1) if $p=O$ and $q=E$ then you have $O+E+E=O=0$, which is impossible

2) if $p=E$ and $q=O$ then you have $E+E+O=O=0$, which is impossible

3) if $p=O$ and $q=O$ then you have $O+O+O=O=0$, which is impossible

4) if $p=E$ and $q=E$ then you have$E+E+E=E=0$. which could be possible ($0$ is an even number), so we treat this case below

Suppose $p=2k$ and $q=2l$ is a solution, then you have that $(p,q)$ is a solution of the equation if and only if $(k,l)$ is a solution, if $k=2e$ and $l=2f$ then you have that $(p,q)$ is a solution if and only if $(e,f)$ is a solution, proceeding in this way you can see that solution, if it exists, must be of the form $(p,q)=(g2^r,h2^s)$ and this also cannot be a solution beacuse this reduces to one of the first 3 cases.

Answer 8

Recall that the square of an odd number is always equivalent to $1$ modulo $8$.

If $a,b,c$ are all odd, then

$$ b^2 - 4 a c \equiv 5 \pmod 8 $$

and thus, $b^2 - 4ac$ cannot be the square of an integer (and thus cannot be the square of a rational number). Therefore the roots are irrational.

Answer 9

$x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} $ or $\dfrac{-b- \sqrt{b^2-4ac}}{2a}$

If $x=\dfrac{p}{q} \implies b^2-4ac=k^2$, and $k$ is odd. ($odd-even=odd$).

Considering $a,b,c$ odd.

$k^2 \equiv 1 \mod 8$

$b^2 \equiv 1 \mod 8$

$4ac \equiv 4 \mod 8$

$b^2-4ac=-3 \mod 8 \implies 5 \mod 8$, a contradiction.(Either of $a$ or $c$ has to be even), since $k^2 \equiv 1 \mod 8$

$\therefore$ You don't get rational roots when all are odd.

Answer 10

Here is a proof that is elementary, i.e. requires no knowledge of modular arithmetic. First we use the AC method to reduce to a monic quadratic, i.e. one with leading coefficient $= 1.$

$$\rm\begin{eqnarray} 0\: =\ f(x)\, = &&\rm\ \, a\,x^2+\,b\,x+c\\ \rm \Rightarrow\ \ 0 = a f(x)\, = &&\rm\! (ax)^2\! + b (ax) + ac\\ \rm \Rightarrow\ \ 0 =\, F(X) = &&\rm\ \ \, X^2 \,+\ \color{#C00}b\,\ X\ \,+ \color{#0A0}{ac},\quad X = ax\end{eqnarray}$$

Suppose $\rm\,f(x_i)=0\,$ for $\rm\:\color{brown}{x_i\in\Bbb Q}.\:$ Then $\rm\,F(X_i)=0\,$ for $\rm\: X_i = a\, x_i\in\Bbb Q.\:$ By the Rational Root Test, the rational roots $\rm\,X_i$ are integers. Since their product $\rm = \color{#0A0}{ac}\:$ is odd, the roots are both odd, so their sum is even. This contradicts: by Vieta, the root sum $\rm\, = -\color{#C00}b\,$ is odd, by hypothesis. Hence $\rm\:\color{brown}{x_i \not\in \Bbb Q}.$

Remark $\ $ This yields a conceptual view of the calculations in Abhijit's answer (which, alas, is incomplete, since it does not justify the reduction from rational to integer roots).

The AC-method generalizes to higher degree polynomials (see the above-linked answer). It is intimately connected to various refinement-based views of unique factorization.

Answer 11

Proof by contradiction, by someone who just started doing proofs, using a similar contradiction proof to "$\sqrt 2$ is irrational". Edit note: You do not need to know mods or virtually anything other than introductory proof by contradiction to do this method.

Let $a,b,c$ be odd. Suppose there is a rational number $p/q$ that can be plugged in for x such that there is a rational solution. Assume $p, q$ have no common factors. (This is where the contradiction will occur!). $0 = ax^2 + bx + c$ has roots $$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$ Plugging in $p/q$ for x gives $$\frac{p}{q} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$ Since $b$ is odd, $b^2$ is the product of odd and odd, meaning $b^2$ is odd. Since $a$ and $c$ are odd, the product of $a$ and $c$, odd and odd, is odd. Therefore $4ac$ is the product of even and odd which is even. So, $b^2 - 4ac$ is odd - even, which is odd. $\sqrt{b^2-4ac}$ must be a square number for there to be a rational solution. Therefore $n^2$ = $\sqrt{b^2-4ac}^2. So, $$n^2$ = ${b^2-4ac}$. We know $b^2-4ac$ is odd, so $n^2$ is odd. If $n^2$ is odd then n is odd, and if n is odd then the entire $\sqrt{b^2-4ac}$ is odd. So, $-b$ $\pm$ $\sqrt{b^2-4ac}$ is odd plus odd, which is even. Almost there, hang in there! 2a is even by definition. If p = -b $\pm$ $\sqrt{b^2-4ac}$ and $q = 2a$, then $p$ is even and $q$ is even. If $p,q$ are even, they are by definition divisible by 2. Therefore, $p,q$ have a common factor. CONTRADICTION!

Let me know if this works:).

Answer 12

Method 1:

Assume $x$ is a root that can be written as $\frac{p}q$, where $ \gcd(p,q) = 1$

$$\frac{ap^2}{q^2} + \frac{bp}{q} + c = 0$$ Multiplying by $q^2$, we obtain: $$ ap^2 + bpq + cq^2 = 0 \tag{1}$$ This maybe looked at as: $$ p(ap + bq) = -cq^2 \tag{2}$$ $$ q(cq + bp) = -ap^2 \tag{3}$$ In (2), $p$ divides the LHS, and thus must divide the RHS, but $ \gcd(p,q) = 1 \implies p | c$. Similarly $ q | a$. Also since $a,c$ are odd, $p,q$ must be odd.

Then, in (1) $ap^2$ is a product of odd numbers and therefore odd. Similarly, every term is odd. Then we have a sum of 3 odd numbers is even ( 0 is even ), which is never possible. Thus by contradiction, there is no rational root

Method 2:

Letting the two roots be $x$ and $y$, written as $\frac{p}{q}$ and $\frac{r}{s}$.

By Rational Root Theorem, we have $ p,r | c$ and also $q,s | a$. Thus $p,q,r, s$ are all odd, as they divide odd integers.

Then by Vietta's Formula, $$ -\frac{b}{a} = x + y = \frac{ps+rq}{qs}$$ $$ bqs = -a(ps+rq)$$ Here, $ps,rq$ are odd and their sum $ps + rq$ is even. Thus the RHS is even but the LHS is odd. Contradiction, and therefore no root is rational

Answer 13

Assume the your root z is rational (= q/r where at at most one of q, r is even)

Re-write the equation with z on the left hand size and 1/z * c/a - b/a on the other. Substitute q/r for z, and put the left hand side into a fraction.

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If neither q or r are even then numerators don't match (q != rc - qb because odd != odd + odd) and if only one is even then the denominator doesn't match (r != qa because odd != even * odd and even != odd * odd)

Therefore, z is not rational.