Polynomial satisfying a relation for all positive integers 
Let $P \in \mathbb{R}[X]$ such that $$P(1)+P(2)+\dots+P(n)=n^5,$$
$\forall n\in \mathbb{N}$. Compute $P\left(\frac{3}{2}\right)$.    

I think that from that relation it is mandatory that $\deg P=5$, but from the given relation we also have that $P(1)+P(2)+...+P(n-1)=(n-1)^5$, so it follows that $P(n)=n^5-(n-1)^5,\forall n\in \mathbb{N}$, so $\deg P=4$.
I know that this only holds for positive integers, but it still seems kind of contradictory to me. Anyway, I don't know if any of my ideas actually help to solve the actual task, so I am looking forward to seeing your ideas.
 A: It is not contradictory. Consider $P(x)=x$ of degree $1$ then
$$P(1)+P(2)+\dots +P(n)=1+2+\dots +n=\frac{n(n+1)}{2}$$
which is of degree $2$. Note also that the sum has not a fixed number of terms, it increases with $n$.
As regards your specific problem, the polynomial $$P(x)-(x^5-(x-1)^5)$$ has infinite zeros, i.e. all the positive integers, so it is the zero polynomial and we may conclude that
$$P(x)=x^5-(x-1)^5.$$
A: In fact we have that
$$
\eqalign{
  & \sum\limits_{k = 1}^n {P(k)}  = n^{\,5} \quad \left| {\;\left( {1 \le } \right)n \in N} \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  P(1) = 1 \hfill \cr 
  \sum\limits_{k = 1}^{n + 1} {P(k)}  - \sum\limits_{k = 1}^n {P(k)}  = P(n + 1) = \left( {n + 1} \right)^{\,5}  - n^{\,5}  \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad P(n) = n^{\,5}  - \left( {n - 1} \right)^{\,5} \quad \left| {\;\left( {1 \le } \right)n \in N} \right. \cr} 
$$
To solve your perplexity about $P(n)$ being of degree $4$ wrt to the sum giving a degree of $5$
consider that the sum of a polynomial with variable upper bound produce the same effect as
the integral: i.e. it raises the degree by $1$.
