Periodicity of algebraic functions I came across this result - An algebraic function cannot be periodic unless it is a constant function. I get the intuition of solving it by writing the function in the form of a polynomial equation, where this function will be one root of the aforementioned equation, and then replace the 'x' in the equation by 'x+T' where T is the period of the function. Then by some further argument, we can simplify things and get to some contradiction that no such T could exist. But I am having difficulty in proving this.
 A: Identity Theorem-
A polynomial $f(x)$ of degree $n$ is identically zero if it vanishes for at least $n+1$ distinct values of $x$.
Proof- Let $\alpha_1,\alpha_2,\alpha_3,...\alpha_n$ be the roots of the $n$th degree polynomial $f(x)$. Then,
$$f(x)=a(x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)$$
Let $\alpha_{n+1}$ be the $n+1$th value for which $f(x)$ vanishes. Then
$$f(\alpha_{n+1})=a(\alpha_{n+1}-\alpha_1)(\alpha_{n+1}-\alpha_2)...(\alpha_{n+1}-\alpha_n)=0$$
$$ \implies a=0\implies f(x)=0$$
Using above result we can say that two polynomials $f(x),g(x)$ of degree $m$ and $n$ respectively (with $m$ less than or equal to $n$) are equal if they have equal values at $n+1$ distinct values of $x$.
Proof- Let $$P(x)=f(x)-g(x)$$
now degree of $P(x)$ is at most $n$ and vanishes for $n+1$ distinct values of $x$. Hence, $P(x)$ is identically zero and thus $f(x)=g(x)$.
Corollary-
The only periodic polynomial function is the constant function.
Proof-   Let $$f(0)=c$$
Now, $$f(T)=f(2T)=...=c$$
Hence, the polynomial function $f(x)$ and $g(x)=c$ take equal values at infinitely distinct values of $x$. Hence, they must be identical. 
Thus, we have, $$f(x)=C$$ for all periodic polynomial functions $f(x)$.

