<s>[INST] "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.
 [/INST] Brainstorming. Given
$$ab=q$$
$$a+b=p+1$$
where $b=a^2+1$, we obtain $$a+\frac{q}{a}=p+1 \Rightarrow a^2-(p+1)a+q=0$$
with (recalling Vieta's)
$$a,b=\frac{p+1\pm\sqrt{(p+1)^2-4q}}{2}$$

One counter-example satisfying all of the above, is $a=\frac{\sqrt{2}-1}{2}$ where $a^2+1=\frac{7}{4}-\frac{\sqrt{2}}{2}$ and $p=\frac{1}{4}\in\mathbb{Q}$, whether $q=\frac{9\sqrt{2} - 11}{8}$ - irrational.

In fact, this reveals many more counter-examples like $a=\frac{k\sqrt{2}}{2}-\frac{1}{2} \Rightarrow a^2+1=\frac{2k^2+5}{4}-\frac{k\sqrt{2}}{2}$ and $p=\frac{2k^2-1}{4}\in\mathbb{Q}$, $q=\frac{(2k^3+7k)\sqrt{2} - 6k^2-5}{8}$ - irrational, for $k$ - integer.

Or even more for $a=\frac{k\sqrt{m}}{2}-\frac{1}{2} \Rightarrow a^2+1=\frac{mk^2+5}{4}-\frac{k\sqrt{m}}{2}$ and $p=\frac{mk^2-1}{4}\in\mathbb{Q}$ ... where $m$ - not a perfect square, covering the $p=1$ ($k=1$, $m=5$) case as well.
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