Polynomial satisfying $p(x)=3^{x}$ for $ x \in \mathbb{N}$ Let $p(x)$ be a polynomial with integer coefficients, which is not constant. Then is this condition possible: $$p(x)=3^{x}$$ whenever $x \in \mathbb{N}$.
Motivation: https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial
 A: The answer depends on the ring (which the question did not specify) where $x$ and $p(x)$ take their values. Polynomials with integer coefficients, and non-negative powers of integers such as $3^n$, can be evaluated in any ring, so the question makes sense in this generality.
In a ring where 2=0 (such as the integers modulo 2), $3^n = 1^n$, and any polynomial with $p(0)=p(1)=1$ is an example.  It should be an interesting problem to determine all rings where there is such a $p(x)$. 
ADDED: Let's consider the situation over an arbitrary commutative ring $R$ with unit.  $Q(x) = p(x+1)- 3p(x)$ is zero (as a function) for all $x$ in the image of $\mathbb{Z}$ in $R$.  If $R$ is of characteristic different from $2$, $\deg Q = \deg P$ (as polynomials).  In characteristic 0 having zeros at all integers is enough to force $Q = 0$ as a polynomial, and there are no examples.  In finite characteristic $n > 1$, we also have $p(n) = 3^n = p(0) = 1$ so that $n | (3^n - 1)$. If $n$ is prime then $n | (3^n - 3)$ so that $n | 2$ and thus $n=2$.  If $n$ is not prime, and $q$ is a prime dividing $n$, then an example in $R$ is an example in $R/q$ which has prime characteristic $q$ and therefore $q=2$.  So the only remaining possibility is a ring where $2^k = 0$, for $k > 1$.  For such rings there are examples because in the binomial expansion of $(1+2)^n = 1 + 2n + 2n(n-1) + 4n(n-1)(n-2)/3 + 2n(n-1)(n-2)(n-3)/3 + \dots$ all terms are congruent modulo $2^k$ to polynomials with integer coefficients, and all terms of degree higher than $2^k - 1$ vanish on the integers modulo $2^k$.
ADDED-2:  one can see that $2^k=0$ more directly by writing $p(n+1)-p(n) = 2p(n)$ (for images of integer $n$ in our ring).  This means that $\Delta^k p(n) = 2^k p(n)$ and for $k > deg(P)$ this will be zero, then set $n=0$ to get $0 = 2^k p(0)= 2^k$.  [Note that this argument applies to $p(n)$ for integer $n$ only, not $p(x)$ as an abstract polynomial or a polynomial evaluated at non-integer values of $x$.]
Conclusion: there are polynomials $p(x)$ such that $p(n)=3^n$ for all $n \geq 0$ (with the polynomial considered as a function on the commutative ring-with-unit $R$) if and only if in this ring, $2^k = 0$ for some positive integer $k$.  
A: $\rm\ p(x+1) = 3\:p(x)\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(x+1) = 3\:p(x)\ \Rightarrow\ p(-1) = 1/3\ $ contra $\rm\ p(x)\in \mathbb Z[x]\:$.
Or, continuing,  $\rm\ p(-n) = 3^{-n}\: \Rightarrow \ p(x)\:p(-x) = 1\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(-x)\:p(x) = 1\ $ contra degree comparison.  This proof works over much more general coefficient rings, e.g. $\mathbb Q\:$.
A: Suppose $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 3^x$ for $x \in \mathbb N$.
Then $a_n (x+1)^n + a_{n-1} (x+1)^{n-1} + \cdots + a_0 = 3 a_n x^n + 3 a_{n-1} x^{n-1} + \cdots + 3 a_0$.
But the coefficient of $x^n$ on the LHS is $a_n$ and on the RHS it is $3 a_n$.
This contradicts the assumption.
A: If so then then the values of $\rm\: q(x) = p(x^2)\:$ are all powers of 3, contradicting the fact that infinitely many primes must occur among the values of a nonconstant $\rm\: p(x) \in \mathbb Z[x]\:$, by way of a simple Euclid-like proof. 
A: Here's another argument (at the other end of the real line!):
Such a polynomial would have to be non-constant, which means that it tends to $\infty$ or $-\infty$ as $x\to-\infty$. But this is impossible, since $3^x \to 0$ as $x\to-\infty$.
A: Suppose that $p(x)$ exists and has degree $n$.  $\lim_{x\to\infty}\frac{p(x)}{x^{n+1}}=0$ and $\lim_{x\to\infty}\frac{3^x}{x^{n+1}}\to\infty$, but $\frac{p(x)}{x^{n+1}}=\frac{3^x}{x^{n+1}}$ for all $x\in\mathbb{N}$, so it is impossible for the first limit to be finite and the second limit to be unbounded.
A: No, a polynomial can't grow exponentially as $x\to\infty$.
A: Yet another proof: if $p$ is a nonzero polynomial with real coefficients
then
$$\lim_{n\to\infty}\frac{p(n+1)}{p(n)}=1$$
which certainly doesn't hold for $p(x)=3^x$.
A: A polynomial of degree $n$ has the property that $p(x+1) - p(x)$ is a polynomial of degree $n-1$.  After taking $n$ finite differences you reach a contradiction.
A: $\rm\ p(x) = 3^x\:$ on $\rm\:\mathbb N\ \Rightarrow\ p(2x) = p(x)^2\:$  on $\rm\:\mathbb N \ \Rightarrow\ p(2x) = p(x)^2\:$ contra degree comparison.
