Complex polynomial P with $P(n)= (-1)^n$ I want to show that there is no polynomial P with complex coefficients such that
$P (n) = (−1)^n$ for all integers n.Does there exist an entire function
with this property ?
Thank you.
 A: To elaborate (slightly) on the comments:  if $P(x)$ were such a polynomial, then both $P(x)+1$ and $P(x)-1$ would have infinitely many zeroes (and hence would be constant).
$cos(\pi x)$ is an entire example. 
A: Ad absurdum, suppose that $P$ is such a polynom. Let $Q$ the polynom with the real part of the coefficient of $P$, id est $P=Q+iT$, where $Q,T$ are real polynoms.
If $\forall n, P (n) = (−1)^n$, then $\forall n, Q (n) = (−1)^n$ (indeed $P(n)=Q(n)+iT(n)$, and $Q(n),T(n)$ are both real because $n$ is real and $Q,T$ are real polynoms, so by identification of the real part of $(-1)^n$, we have $Q (n) = (−1)^n$), then $\forall n, \exists a \in (n, n+1)$ such as $Q(a)=0$ (consequence of the intermediate value theorem). So $Q$ has an infinity of root, so $Q=0$, but $Q(0)=1$, it is absurd. 
So there is no such polynom.
The example for an entire function is given in the comment section :).
A: If $P$ is polynomial, then $|P(x)|\to \infty$ as $|x| \to \infty,$ unless $P$ is constant. Thus if $(a_n)$ is a bounded sequence with more that one distinct term, then there is no polynomial $p$ such that $p(n) = a_n$ for all $n.$
