How do I prove that if $p$ and $q$ are odd numbers then $x$ is not a rational number in $x^2 + 2px +2q=0$?

So I don't know a whole bunch about quadratic functions behaviour but I know that we can get the solution for $x$ with the quadratic formula. I tried comparing the general formula with a fraction in which either $m$ or $n$ is an irrational number.

\begin{equation*} \qquad \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{m}{n} \end{equation*}

We know that n is not the irrational number since a equals $1$ . Therefore the upper part of the quadratic formula should equal an irrational number. I'm stuck on that part though, which function property could I use to prove this?

Note: I'm new to this site and I am learning to correctly use the features of stack exchange. If there is something I can improve in I'll be more than happy to take some constructive criticism


marked as duplicate by Parcly Taxel, Joey Zou, Claude Leibovici, Michael Hoppe, Community Sep 15 '16 at 23:15

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    $\begingroup$ Oh wait? I answered this question here! $\endgroup$ – Parcly Taxel Sep 15 '16 at 6:31
  • $\begingroup$ It's going to be irrational if and only if $\sqrt {b^2-4ac} $ is irrational. Express this in terms of p a q and see what happens. If p and q are odd, you should get that this is irrational. $\endgroup$ – fleablood Sep 15 '16 at 7:06

The discrimiant is $$\sqrt{(2p)^2-8q}=2\sqrt{p^2-2q}$$

We need $p^2-2q=r^2$ where $r$ is an integer

As $p$ is odd, so will be $r^2,r$

Finally $p^2-r^2=(p+r)(p-r)$ is a multiple of $2\cdot4$ unlike $2q$


$$x^2 + 2px + 2q = 0$$

Suppose $u$ is a rational root of the above equation. By the rational root theorem, u must be an integer. Since $0$ is an even number, $u$ must be an even number, say $u = 2v$ for some integer $v$. Substituting and simplifying, we find

$$2v^2 + 2pv + q = 0$$

But $2v^2 + 2pv + q$ is an odd integer and $0$ is an even integer.

By contradiction, there are no rational roots.


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