Let $f$ be an entire function, and let $\alpha, \beta \in \mathbb{C}$ such that $$f(z+\alpha) = f(z) \quad f(z+\beta) = f(z) \quad \forall z \in \mathbb{C}$$ Then if $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$, $\{n\alpha +m\beta: n,m\in \mathbb{Z}\}$ is dense in $\mathbb{R}$. Prove that $f$ is constant.
I know I have to come up with some kind of sequence converging to any point on $\mathbb{R}$ but I dont know how to do it, and what it has to do with the fact that $\alpha $ and $\beta$ are linearly independent. The second part is easier, I can just use Liouville's theorem, as f is bounded over the real line because $f(0) = f(n \alpha)$ and f is bounded at 0, but I don't know if I can prove that f is bounded just by looking at if on the real line.
Any help will be appreciated.
Case 1 $\alpha, \beta$ are linearly dependent over $\mathbb R$. This means $\beta=s\alpha$ for some $s \in \mathbb R$.
Show first that
$$ \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}=0$$
Indeed, assume by contradiction that $$ a= \mbox{ inf } \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}>0 $$
First show that $a \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$.
Note that otherwise you could find two elements $b,c \in \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ with $$ a<b<c <\frac{3a}{2} \,. $$ But then $c-b \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ and $c-b <a$ contradicting the definition of $\mbox{inf}$.
Next, since $a \in \mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\} $ it follows that $$ \mathbb Z a \alpha \subseteq \{n\alpha +m\beta: n,m\in \mathbb{Z}\} \,. $$
Complete the contradiction by observing that the above inclusion must be equality, as otherwise you can create an element $d \in\mbox{ inf } \{\frac{n\alpha +m\beta}{\alpha}:\frac{n\alpha +m\beta}{\alpha}>0; n,m\in \mathbb{Z}\}$ with $$ 0 < d< a $$ which is not possible.
You can now reach the contradiction: $$ \mathbb Z a \alpha = \{n\alpha +m\beta: n,m\in \mathbb{Z}\} $$ gives that $$ \alpha =ka \alpha \\ \beta=la \alpha$$ The first relation gives $a \in \mathbb Q$, and then the second contradicts the linear independence over $\mathbb Q$.
Next, show that $\{n\alpha +m\beta: n,m\in \mathbb{Z}\}$ If $(e,f)$ is dense in the line $\mathbb R \alpha$ . Indeed, if $(e,f) \subseteq \mathbb R$ is any interval, by the above there exists some $m,n$
$$ 0<\frac{n\alpha +m\beta}{\alpha}< f-e $$ Pick the smallest $k \in \mathbb Z$ such that $k\frac{n\alpha +m\beta}{\alpha}>e$ and show that $k(n \alpha+m\beta) \in (e\alpha ,f \alpha)$.
FInally, since $f$ is constant on $\{ n\alpha +m\beta : m,n \in \mathbb Z\}$ and continuous, it is constant on its closure which is $\mathbb R \alpha$. The claim follows.
Case 2 $\alpha, \beta$ are linearly independent over $\mathbb R$.
In this case you can show that there exists some $R>0$ such that for each $z \in \mathbb C$ there exists $m,n \in \mathbb Z$ such that $$ |z-m\alpha-n \beta| \leq R \,. $$
Since $f$ is bounded on $\{ w : |w| \leq R \}$, the above implies that $f$ is bounded on $\mathbb C$.