Question. Let be an entire non-constant function that satisfies the functional equation
for all . Show that .
for some polynomial . Therefore,
for all . From the fact that
Put and , we obtain
which implies . From the identity put we then deduce that
Note: I think I should post some applications of the Hadamard’s Theorem.
Corrigendum to the previous proof. Assume by contradiction, then takes all values except possibly some . Also notethat . So we can assume which means that . It then does take the value , and hence takes the value . The proof is now complete.
I thank Xu Wei Biao for pointing out a mistake in the previous solution.