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Hartshorne book "Algebraic geometry" Proposition I.7.3 (b),

Proposition I.7.3.(b) : If $f: \mathbb Z \rightarrow \mathbb Z$ is any function, and $Q$ is a numerical polynomial such that $\Delta f=f(n+1)-f(n)=Q(n)$ for all $n\gg 0$, then there exists a numerical polynomial $P$ such that $f(n)=P(n)$ for all $n\gg 0$.

In proof of this proposition, If $\Delta(f-P)=0 \text{ for } n\gg 0$, then $(f-P)(n)$ is constant for $n\gg 0$ and so $f(n)=P(n)$ for $n\gg 0$.

But I can't understand this proof. May be I see more in detail.?

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    Let $g: \mathbb Z \rightarrow \mathbb Z$ be any function. If $\Delta(g)=0 \text{ for } n\gg 0$, then $g(n) = g(n+1) = g(n+2)=\cdots$ for $n\gg 0$. So $g(n)$ is constant for $n\gg 0$.2012-09-06

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