Is it 'rare' that $a$ and $a+1$ are conjugate (= have the same minimal polynomial)?

Question

Let $a \in \bar{k}-k$, $k$ is a field of characteristic zero and $\bar{k}$ is an algebraic closure of $k$.

Denote the minimal polynomial of $a$ by $m_a=m_a(t) \in k[t]$.

Is it 'rare' that $m_a=m_{a+1}$? In other words, is it rare that $a$ and $a+1$ are conjugate?

For example, $k=\mathbb{Q}$ and $a=\sqrt{2}$. Then $m_a=t^2-2 \neq m_{a+1}$.

Any hints are welcome!

Answer 1

If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $\deg f>0$), it follows tat $g=f$. So as a function on $\Bbb Z$, $f$ is periodic, hence bounded, hence constant ...

Answer 2

Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?