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How would I show the following.

For all odd integers a,b, and c if z is a solution of $ax^2+bx+c=0$ the z is not rational.

How would I show this? This is what I did

Let a be an odd integer then for a integer k it has the form $a=2k+1$

then $b=2t+1$ for if c integer.

then $c=2r+1$ if s is an integer then I do

$(2k+1)(x)^2+2t+1(x)+(2r+1)=0$

But I find my self stuck on what to do any hint are appreciated.

Bill Dubuque
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Fernando Martinez
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4 Answers4

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Hint: Show that the discriminant is not a perfect square. Note that $b^2\equiv 1\pmod{8}$, and $4ac\equiv 4\pmod{8}$, and therefore $b^2-4ac\equiv 5\pmod{8}$.

Another way: If there are rational roots, then the quadratic factors as $(px+q)(rx+s)$, where $p,q, r, s$ are integers. Because of the oddness of $a$ and $c$, $p,q,r,s$ are all odd. But then the coefficient of $x$ in the product is even.

André Nicolas
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  • I see it is not rational because the discriminant is not a perfect square then when you divide it by 2 in the quadratic – Fernando Martinez Dec 07 '13 at 22:01
  • If we use the quadratic formula, the roots are $\frac{-b\pm\sqrt{D}}{2a}$. If this is rational, then $\sqrt{}$ is rational. But it isn't. Note the alternate approach that uses less machinery, though it too leaves detail to be filled in. – André Nicolas Dec 07 '13 at 22:06
  • I see i think I get the alternative way I see now that because a and c are odds the factors are odd but I a not sure is the coefficient of x in the product 0? – Fernando Martinez Dec 07 '13 at 22:10
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    The coefficient of $x$ is $ps+qr$. Since $ps$ and $qr$ are odd, their sum is even. I prefer the version of the parity argument given by Ian Mateus. – André Nicolas Dec 07 '13 at 22:17
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Hint: set up $x=m/n$, $\gcd(m,n)=1$ and look modulo $2$. You'll get $$am^2+bn+cn^2\equiv0\pmod2.$$ We have three cases on the parity of the terms. What happens?

It is particularly easy to do it for integer roots. Suppose $ax^2+bx+c=0$. If $x$ is even, then $ax^2+bx+c=x(ax+b)+c$ is odd, so it can't be zero. Similarly, if $x$ is odd, then we have the sum of three odd numbers, which is again odd. You can adapt this approach for rationals, as stated above.

Ian Mateus
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Have you looked at the contrapositive yet?

If both roots are rational, then at least one of $a$, $b$, or $c$ are even.

Or equivalently, for example,

If both roots are rational and $a$ and $c$ are odd, then $b$ is even.

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If $\rm\ z = m/n\,$ in lowest terms, Rational Root Test $\rm\,\Rightarrow m\mid c,\ n\mid a,\ $ so $\rm\,\ c,a\,\ odd\Rightarrow m,n\,\ odd,$

hence $\rm\,\ \color{#c00}0 = n^2 (a z^2\! + b z + c) = a m^2\! + b mn + c n^2 = odd + odd + odd = \color{#c00}{odd}\ \Rightarrow\Leftarrow\ $ QED

Remark $\ $ The idea of using parity to exclude the existence of roots generalizes, namely

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

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

The Parity Root Test generalizes to any ring with a sense of parity, e.g. the Gaussian integers $\rm\: a + b\,{\it i}\ $ for integers $\rm\:a,b.\:$ For much further discussion see this post and also these related posts.

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Bill Dubuque
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