"If $a+a^3$ is irrational, then $a+a^2$ is also irrational". Given $a\in\mathbb{R}$, define $p=a+a^2$ and $q=a+a^3$. Show that the following statement is false: 
"If $q$ is irrational, then $p$ is also irrational".
My approach: I just was lucky to find a counterexample setting $p=1$. Then, I found out that for both $a$ that satisfy the equation $q$ is irrational.
I wonder if someone here could come up with a more neat solution.

$a^2 +a - p=0$ so $a = \frac{-1\pm\sqrt{1+4p}}2$ and so
$q = (\frac{-1\pm\sqrt{1+4p}}2)^3 + ((\frac{-1\pm\sqrt{1+4p}}2))$
To get a counter example $p$ needs to be rational $q$ does not.
$\sqrt{1+4p}$ will only be rational if $1+4p$ is perfect rational square.   It is trivial easy to choose rational $p$ where $4p+1$ is not perfect rational square.
If $\sqrt{1+4p}$ is rational then
$q = (\frac{-1\pm\sqrt{1+4p}}2)^3 + ((\frac{-1\pm\sqrt{1+4p}}2))=$
$r_1 + r_2*\sqrt{1+4p} + r_3*\sqrt{1+4p}^2 + r_4*\sqrt{1+4p}^3=$
$r_1 + r_2*\sqrt{1+4p} + r_3*(1+4p) + r_4*(1+4p)\sqrt{1+4p}=$
$s_1 + s_2\sqrt{1+4p}$ for some $r_1, r_2, r_3, s_1, s_2$.
So long as $s_2\ne 0$ and $1+4p$ is not a perfect square, then $q$ will be irrational.
And $s_2 =0$ is trivially easy to avoid[1].
So there was nothing particularly "lucky" about choosing $p=1$.
[1] $s_2 = r_2 + r_4(1+4p)$
$r_2 = 3(-\frac 12)^2(\pm \frac 12)+1=\pm \frac 38 + 1$ and $r_4=(\pm \frac 12)^3=\pm \frac 18$
So $s_2 = \pm \frac 12 +1  \pm \frac 12p$.
So there are precisely two values of $p$ to avoid.
