**Question.** Let be an entire non-constant function that satisfies the functional equation

for all . Show that .

**Solution.** Assume by contradiction, then W.L.O.G. by the Picard’s Theorem we can assume that misses . By the Hadamard’s Theorem,

for some polynomial . Therefore,

for all . From the fact that

we get

which yields

.

Put and , we obtain

.

Hence

which implies . From the identity put we then deduce that

a contradiction.

*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.

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By the Hadamard’s Theorem, how can I say f(z)-a= e^p(z)????

Comment by scsdxc — May 4, 2016 @ 1:15

I think there is a big calculation mistake after the line “which yields” and after calculation I can’t show a=1/2..please correct this mistake

Comment by RUHUL ALI KHAN — May 4, 2016 @ 2:27

Thank you. To save our time, please let me know the mistake so that we can fix asap :).

Comment by Ngô Quốc Anh — May 4, 2016 @ 2:36

You are right, there was missing a minus sign in front of :(.

Comment by Ngô Quốc Anh — May 4, 2016 @ 2:43

inspite of that I can’t show a=1/2

Comment by RUHUL ALI KHAN — May 4, 2016 @ 15:57