If at least one solution $r_1, r_2$ to $ax^2 + bx +c$ is rational, then prove that at least one of $a, b, c$ must be even. I have been trying to prove the contrapositive but have been facing trouble. Can anyone help me?
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1$(x+1)(x+\pi)=x^2 + (1+\pi)x + \pi$? – Calvin Khor Sep 16 '19 at 01:03
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Do you want $a$, $b$, $c\in\Bbb Z$? Recall that $x^2+x+1$ is irreducible modulo $2$. – Angina Seng Sep 16 '19 at 01:06
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@LordSharktheUnknown That's the most reasonable assumption imo. – Rushabh Mehta Sep 16 '19 at 01:08
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$a, b,c $ are integers – Albert Chung Sep 16 '19 at 01:20
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The quadratic equation is that $r_1, r_2 = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$. And that number is rational if and only if $b^2 - 4ac$ is a perfect square. So prove if $b^2 -4ac$ is a perfect square then one of $a,b,c$ is even. – fleablood Sep 16 '19 at 02:12
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4Exact duplicate of Roots of $,ax^2+bx+c,$ are irrational if $a,b,c$ are odd integers – Bill Dubuque Dec 21 '19 at 05:44
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Using the quadratic formula, you get
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \tag{1}\label{eq1}$$
Since $a,b,c$ are integers, for $r_1$ or $r_2$ to be rational requires the discriminant to be a perfect square, i.e., there's an integer $d$ such that
$$b^2 - 4ac = d^2 \tag{2}\label{eq2}$$
If $a,b,c$ are all odd integers, then $b^2 \equiv 1 \pmod 8$ and $4ac \equiv 4 \pmod 8$, so $d^2 = b^2 - 4ac \equiv 1 - 4 \equiv 5 \pmod 8$, but squares only have congruences of $0,1,4$ modulo $8$. Thus, the assumption must be false, i.e., at least one of $a,b,c$ is even.
John Omielan
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@RyanTong As explained in the Definition of congruence relation section of Wikipedia's "Modular arithmetic" article, $a \equiv b \pmod n$ means $a$ and $b$ have the same remainder when divided by $n$. As such, for some integer $k$, you have $a = kn + b$. For example, in my first case of using modulos, I wrote if $b$ is odd, then $b^2 \equiv 1 \pmod 8$. This is because $1^2 = 1$, $3^2 = 9 = 8 + 1$, $5^2 = 25 = 3\times 8 + 1$, etc. Also, $4ac \equiv 4 \pmod 8$ as odd mult. of $4$ have rem. $4$ when div. by $8$. – John Omielan Sep 16 '19 at 15:33
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If $a$, $b$ and $c$ are integers, then for $ax^2+bx+c$ to have a rational root, the polynomial must factor: $$ax^2+bx+c=(rx+t)(ux+v)$$ where $r,\ldots,v$ are integers. As $a$ and $c$ are odd, all the $r,\ldots,u$ must be odd. But then what about $b=rv+tu$?
Angina Seng
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