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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.

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  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Jul 31, 2019 at 1:45

2 Answers 2

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

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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 $1$.

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