Polynomial equation 
Is it true that if $P \in \mathbb{Z}[X]$, then there exists $Q \in \mathbb{Z}[X]$ such that $P(x)=Q(x+1)-Q(x)$?
Can we generalize to other rings than $\mathbb{Z}$? 

I came up with this by looking at the identity $$\sum_{j=1}^{n}j(j+1)\ldots(j+k-1)=\frac{j(j+1)\ldots(j+k)}{k+1}$$ for any $k \geq 1.$ 
 A: It's sufficient to show for any $n \in \mathbb N$, exists $Q(x) \in \mathbb Z[x]$ such that  $Q(x+1)-Q(x)=x^n$.
If $Q(x)=a_{n+1}x^{n+1}+a_nx^n+...+a_1x$ then $Q(x+1)-Q(x)=$ 
$$a_{n+1}(x+1)^{n+1}+a_n(x+1)^n+...+a_1(x+1)-a_{n+1}x^{n+1}-a_nx^n-...-a_1x =   
\binom{n+1}{1}a_{n+1}x^n+(\binom{n+1}{2}a_{n+1}+\binom{n}{1}a_n)x^{n-1}+...+(a_{n+1}+a_n+...+a_1) =x^n $$
Now $a_{n+1}=\frac{1}{n+1}$ and $a_{n+1-i}$ is obtaining from $a_{n+1} , a_n ,...,a_{n+2-i}$ .
A: As stated, no -- if you want to write $P(x) = x$ as $Q(x+1)-Q(x)$, you would need
$$Q(x) = \frac{1}{2}x^2 - \frac{1}{2}x $$
which is in $\mathbb{Q}[x]$ but not $\mathbb{Z}[x]$. However if you allow $P,Q \in \mathbb{Q}[x]$ the answer is yes, such a $Q$ always exists. To see this, construct the desired $Q$ from monomials by induction on the degree.
If $P(x) = c \cdot x^0$, $Q(x) = c \cdot x$. If $P(x) = c \cdot x$, 
$$ Q(x) = \frac{c}{2} \left( x^2 - x \right) $$
In general, if $P(x) = c \cdot x^n$, then 
$$ Q(x) = \frac{c}{n+1} \left( x^{n+1} + (\textrm{lower order terms}) \right) $$
where the lower order terms compensate for the fact that using $Q(x) = \frac{c}{n+1} x^{n+1}$ as your first pass will give you 
$$Q(x+1) - Q(x) = c \cdot x^n + (\textrm{lower order terms})$$
Induction guarantees that the lower order terms for $Q$ exist to give you what you need. They aren't too hard to write out explicitly.
Now if you have any polynomial $P(x)$, find the correct term for each term in $P$, and sum to get your $Q$.
This technique requires closure under multiplicative inverses of $\mathbb{Z}$, so seems like it only works in $R[x]$ where $R$ is a ring containing a field.
