"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: I want to add the following more general result/method.
Observation. Assume that $f(x)$ and $g(x)$ are polynomials with rational coefficients and respective degrees $m$ and $n$. If $\gcd(m,n)=1$ then for all real numbers $\rho$ the values $f(\rho)$ and $g(\rho)$ can both be rational if and only if $\rho$ is rational.
Proof. Consider the field of rational functions $K=\Bbb{Q}(x)$. It has subfields
$F_1=\Bbb{Q}(f)$, $F_2=\Bbb{Q}(g)$ and $F_3=\Bbb{Q}(f,g)$. We immediate see that 


*

*$F_1\subseteq F_3$ and $F_2\subseteq F_3$.

*$[K:F_1]=m$ and $[K:F_2]=n$.

*By the tower law $[K:F_3]$ must be a factor of both $m$ and $n$, so $[K:F_3]=1$, and therefore $K=F_3$.


This means that $x\in\Bbb{Q}(f,g)$, so there exists bivariate polynomials $a(X,Y), b(X,Y)\in\Bbb{Q}[X,Y]$ such that 
$$
x=\frac{a(f,g)}{b(f,g)}.
$$
The claim follows from this by plugging in $x=\rho$. We were given that $f(\rho)$ and $g(\rho)$ are rational. Consequently so are $a(f(\rho),g(\rho))$ and $b(f(\rho),g(\rho))$ and hence also their ratio $=\rho$. QED
This has the following
Corollary. If $f(x)$ and $g(x)$ are monic polynomials with integer coefficients and respective degrees $m,n>1$ such that $\gcd(m,n)=1$, then there exists a real nuber $\rho$ such that $f(\rho)$ is rational but $g(\rho)$ is not.
Proof. Because $f(x)$ has degree $\ge2$, there exists an integer $m$ such that $f(x)=m$ has a real solution $x=\rho$, but it has no solutions $x\in\Bbb{Z}$. This is basically because asymptotically $|f(x)|$ is growing faster than linearly when $|x|\to\infty$, so the image $f(\Bbb{Z})$ cannot be all of $\Bbb{Z}$. As we assumed $f(x)$ to be monic the rational root test implies that any rational solution of $f(x)=m$ would be an integer. As that is not the case, we can conclude that $\rho$ is irrational.
If $g(\rho)$ were rational then, by the Observation, $\rho$ would have to be rational also. Therefore $g(\rho)$ is irrational. QED

I'm fairly sure that the assumption of $f$ and $g$ being monic is not necessary. For the argument to work we only need to locate a rational number $q$ such that $f(x)=q$ has an irrational solution $\rho$. I don't have the time to think of an argument showing that such a $q$ exists. So I bailed out by assuming that $f$ and $g$ are monic, when the above argument is available.
A: Hint $ $ A counterexample has $q$ irrational but $p$ rational. Substitute $\,a^2 = p-a\,$ into $\,q = a^3+a\,$ to get $\,q = (p\!+\!2)\,a - p,\,$ so  $\,q\,$ is irrational iff $a$ is, if $\,\color{#c00}{p\neq -2}.\,$ So it suffices to find a rational $p\neq -2$ so that $\,a^2+a-p\,$ has an irrational root $\,a,$ which is easy, e.g. your $\,p=1.$
Remark $ $ This works more generally when $f(a)\bmod g(a) = c_1(p)\, a + c_0(p)\,$ has $\rm\color{#c00}{degree\ one}$. Above $\,f(a) = a^3+a,\,$ $\,g(a) = a^2+a-p,\,$ and $\, f\bmod g = (p\!+\!2)\,a - p.$
A: We just need to disprove

If $p=a+a^2$ is rational, then $q=a+a^3$ is rational too.

This is clearly false since by assuming $a+a^2=2m+1$ we have that $a$ is a quadratic irrational ($x^2+x-(2m+1)$ is irreducible in $\mathbb{F}_2$, hence in $\mathbb{Q}$) and 
$$ q = a+a(2m+1-a) = (2m+2)a-a^2 = (2m+3)a-(2m+1) $$
is a quadratic irrational too.
A: 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.
A: $a^2 +a - p=0$ 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.
