Finding a basis for a certain vector space of periodic polynomials I am having a little bit of trouble solving an homework question.
I found that $S={ p(x) \in R_4[x]} \big| p(x)=p(x-1) $ is a vector space.
Now I need to find some set, K that holds ${span(k)=S}$
My idea was to substitute: $ a+bx+cx^{2}+dx^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}$
But here I get something that seems a little bit weird, I'd like to look if I have any mistakes that I did not notice. and if I had mistakes, how to make my answer right.
I calculated the right side of the equation and I got to  :
$a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}=a+bx-b+cx^{2}-2cx+c+dx^{3}-3dx^{2}+3dx-d+ex^{4}-4ex^{3}+6ex^{2}-4ex+e=
(a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4}
 $
From here,
$
(a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}\Rightarrow $
$(-b+c-d+e)+(-2c+3d-4e)x+(-3d+6e)x^{2}+(d-4e)x^{3}=0+0x+0x^{2}+0x^{3}+0x^{4}$
And the solution to this equation is only whan $a=b=c=d=e=0$.
I think that I am maybe not in the right direction.
Do you have any Idea how to find a set $K\Rightarrow span(K)=S$ ?
Thanks in advance. 
 A: Two points about your question.


*

*After you made $a$ drop out of the equation (to obtain the equation that starts with $(-b+c-d+e)\cdots$), you nevertheless concluded that $a=0$ (among other things). This cannot be right, you can never get a conclusion about $a$ from an equation in which it is absent.

*Supposing your conclusion $a=b=c=d=e=0$ would have been justified, then you would have arrived at the conclusion that your subspace contains only the polynomial $0$. This is wrong, but it is not anything contradictory in itself. In any case a subspace requires a number of basis vectors equal to the number of parameters that can be freely chosen. If you've got no such parameters, this would mean the basis has $0$ elements; it is the empty set ($K=\emptyset$).
In fact all constant polynomials (not just $0$) are solutions to your problem, so there is $1$ parameter ($a$) that can be freely chosen. So your subspace requires one basis vector (any nonzero constant polynomial will do for this purpose; the most obvious choice is the constant$~1$). In general, to pass from the description of a subspace in terms of parameters (obtained by solving a linear homogeneous system, as you did), select each free parameter in turn, make it$~1$ while making any other free parameter$~0$, and thus obtain one basis vector of the subspace for each free parameter.
A: Your method is correct, but the conclusion is not---the resulting system imposes no constraints on $a$, so the general solution is $p(x) = a$, and this is spanned by, e.g., the polynomial $p(x) = 1$.
Here's a more conceptual and less computational way to see this: Since any such p is by definition periodic with period 1, all of the values in its image $p(R)$ are achieved on the interval (for example) $[0,1]$. Now, p is a continuous, so $p(R)=p([0,1])$ is compact and hence bounded. But, the only bounded polynomials are constants, and all constant polynomials $p(x)=a$ satisfy the periodicity condition, so $S = \{p(x) = a : \text{$a$ constant}\}$.
A: Suppose there exists a non-constant polynomial $p \in S$ of degree $d \geq 1$. I will get a contradiction, so that $S$ must contain only constant polynomials.
Write $p=\sum_{i=0}^d a_ix^i$ with $a_d \neq 0$ (because the degree is $d$). Then $p(x) = p(x-1)$ is equivalent to
$$\sum_{i=0}^d a_ix^i - \sum_{i=0}^d a_i(x-1)^i=0$$
but this implies that
$$da_d x^{d-1} + (\mbox{ other terms of degree at most $d-2$ }) =0$$
so $a_d=0$, and we get a contradiction.
Note that this argument works in any degree, and in any integral domain.
