suppose that$ \alpha$ and $ \alpha^3+1$ are two root of $ f$ and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3+1$ is also a root of $ f$ we knoew $ (x^3-\alpha^3)|f(x^3+1)$ so $ \alpha$ is a root of $ f(x)$ but we know that $ f(x)$ is minimal polynomial that contains $ \alpha$ as a root so we have $ f(x)|f(x^3+1)$ so if we put $ \beta$ we will see that $ \beta^3+1$ is a root. if we put $ \beta$ with norm greater than $ \frac{3}{2}$ we have $ |\beta^3+1|>|\beta^3|-1>|\beta|$ because root of $ g(x)=x^3-x-1$ with maximum norm is not greater than $ \frac{3}{2}$ so we reach unlimited root for $ f$ which contradicts.

I think I have the same solution as sumita, but I'll write a bit more clearly. I will use the following Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $alpha$ (up to a constant factor). Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$. We also need to use Lemma: If $r \in C$ and $|r|>1$ then $|r^3+1|>|r|$ Let $r_1 \in C$ denote the root of $f$ with norm larger than $3/2$ (i.e $|r_1|>3/2$). By the above lemma, $r_1^3+1$ is a root of $f$ distinct from $r_1$. $(r_1^3+1)^3+1$ is yet another distinct root of $f$, and so on. Thus $f$ has infinitely many distinct roots, so it is the zero polynomial, contradiction.

sumita wrote: suppose that$ \alpha$ and $ \alpha^3+1$ are two root of $ f$ and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3+1$ is also a root of $ f$ I don't understand this part ,can anyone help me?thanks!

Stranger8 wrote: sumita wrote: suppose that$ \alpha$ and $ \alpha^3+1$ are two root of $ f$ and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3+1$ is also a root of $ f$] I don't understand this part ,can anyone help me?thanks! He's trying to prove the statement by contradiction, so he supposes $\alpha$ and $\alpha^3+1$ are both roots. He then says he will prove that if $b$ is any root of $f$, then $b^3+1$ is also a root.

viperstrike wrote: Stranger8 wrote: sumita wrote: suppose that$ \alpha$ and $ \alpha^3+1$ are two root of $ f$ and also suppose $ \beta$is also a root we can show for each $ \beta$,$ \beta^3+1$ is also a root of $ f$] I don't understand this part ,can anyone help me?thanks! He's trying to prove the statement by contradiction, so he supposes $\alpha$ and $\alpha^3+1$ are both roots. He then says he will prove that if $b$ is any root of $f$, then $b^3+1$ is also a root. But how to prove if $b$ is any root of $f$, then $b^3+1$ is also a root. I don't know this sorry can you help me please?

He says it here: we knoew $ (x^3-\alpha^3)|f(x^3+1)$ so $ \alpha$ is a root of $ f(x)$ but we know that $ f(x)$ is minimal polynomial that contains $ \alpha$ as a root so we have $ f(x)|f(x^3+1)$ so if we put $ \beta$ we will see that $ \beta^3+1$ is a root. I also say it here: viperstrike wrote: Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $\alpha$ (up to a constant factor). Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$. .

Oh I get it now, I didn't know how to use "irreducible "condition , so I didn't understand your solution,now I realize the importance of minimal polynomial which helps me understand it well,thank you very much!

viperstrike wrote: I think I have the same solution as sumita, but I'll write a bit more clearly. I will use the following Lemma: If $f \in Z[x]$ is irreducible and $\alpha \in C$ s.t $f(\alpha)=0$, then $f$ is the minimal polynomial of $\alpha$ (up to a constant factor). Suppose $\alpha$ and $\alpha^3+1$ are both roots of $f$ for some $\alpha \in C$. Let $g(x)=f(x^3+1)$. Since $g(\alpha)=0$ and $g \in Z[x]$, we have $f|g$. Therefore if $f(r)=0$ then $g(r)=f(r^3+1)=0$. We also need to use Lemma: If $r \in C$ and $|r|>1$ then $|r^3+1|>|r|$ Let $r_1 \in C$ denote the root of $f$ with norm larger than $3/2$ (i.e $|r_1|>3/2$). By the above lemma, $r_1^3+1$ is a root of $f$ distinct from $r_1$. $(r_1^3+1)^3+1$ is yet another distinct root of $f$, and so on. Thus $f$ has infinitely many distinct roots, so it is the zero polynomial, contradiction.