# 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.

• Even if you assumed $\deg P(x)=5$, you have a number of values to interpolate to determine the six required coefficients. It would turn out a posteriori that the coefficient of $x^5$ is zero. The summation here acts much like an integration, raising the degree of $P$ by one. Jul 31, 2019 at 1:45

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 $$P(x)=x^5-(x-1)^5.$$
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 $$1$$.