How would one prove that a function is a polynomial? I can't seem to find anything about this on the internet. I would like to know if there are any unique properties that only polynomials can satisfy. Given properties of a function, can these properties be used to prove that a function is a polynomial. I would like to do this without actually constructing the polynomial. What I mean is something like this: One can show that the sine function is not a polynomial without deriving an actual formula. You just use the property that a polynomial has finite degree.


EDIT: My function is only defined on the positive integers. So I am not sure how to use the ideas of complex analysis.



What conditions imply that a function over $\mathbb{Z}$ is a polynomial?
This is the same answer as given several times already. It is prefaced with a few facts on Taylor series to make it seem familiar. First recall that a real function $f(x)$ well enough behaved at $x=0$ has a Taylor series $$f(x)=\sum\_0^{\infty}a\_k\frac{x^k}{k!} $$ valid in some interval. And 



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> f(x) is a polynomial if there is an $K$ with $a\_k=0$ when $k \gt K$ (and the interval of validity is the entire real line.)
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Similarly, any real function $f(n)$ defined on $\mathbb{N}$ (a.k.a. a sequence) has a unique expansion $$f(n)=\sum\_0^{\infty}a\_k\frac{(n)\_k}{k!}.$$


valid on all the domain. And



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> $f$ is a polynomial (on $\mathbb{N}$) exactly if there is an $K$ with $a\_k=0$ when $k \gt K.$ 
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This leaves several things to explain. 


For the Taylor series, $a\_k=(D^kf)\ (0)$ where $D=\frac{d}{dx}$ is the differential operator which sends $f(x)$ to the derivative $f'(x)$. 


For sequences on $\mathbb{N}$, $a\_k=(\Delta^kf)\ (0)$ where $\Delta$ is the difference operator which sends the sequence 


$f(0),f(1),f(2),f(3),\cdots$ 


to the sequence of differences 


$f(1)-f(0),f(2)-f(1),f(3)-f(2),f(4)-f(3)\cdots$


Also, $(n)\_k=n(n-1)(n-2)\cdots(n-k+1)$ is the falling factorial. 
Of course the question concerned $\mathbb{Z}$ rather than $\mathbb{N}.$ 



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> For the domain $\mathbb{Z},$ we can first restrict to $\mathbb{N}$, determine if we have a polynomial and, if so, then check that the expansion is valid on all of $\mathbb{Z}.$
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The expressions $\frac{(n)\_k}{k!}$ were used point out the relation to Taylor series. Note, however, that $\frac{(n)\_k}{k!}=\binom{n}{k}$ is a binomial coefficient. 


Examination of these sequences $\binom{n}{k}$ make it clear that they constitute a basis (of sorts) for the space of sequence defined on $\mathbb{N}.$ The first few are


$$\begin{array}{cccccccc}
 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \cdots \\
 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\cdots\\
 0 & 0 & 1 & 3 & 6 & 10 & 15 & 21\cdots \\
 0 & 0 & 0 & 1 & 4 & 10 & 20 & 35 \cdots\\
 0 & 0 & 0 & 0 & 1 & 5 & 15 & 35\cdots 
\end{array}$$
Clearly there is a unique linear combination of these sequences for any target sequence. Although the sum is potentially infinite, it is finite for each fixed value of $n.$


Furthermore, $f$ takes integer values on all of $\mathbb{N}$ exactly if the $a\_k$ are all integers.


