**Problem: ** Suppose . If is a polynomial of for every fixed and a polynomial of for every fixed . Then must be a polynomial.

*Proof:* Since is holomorphic on the , we have the power series expansion

where is a holomorphic function on for any . Suppose there exists a sequence with satisfying . Since the set of roots of are countable in , then all the union of these sets are also countable. Therefore, such that for all . Then can never be a polynomial. So

where every is polynomial of . Repeating the above proof we get this is also a finite sum, which means is a polynomial.

**Remark:** Exercise from Shabat, 1.10. Idea comes from Hanlong Fang.

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