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I was asked a while ago to prove there is no polynomial $P$ in $\mathbb R$ such that $P(i)=f_i$ for all $i\geq0$. I tried to get a proof as slick as possible and here's what I got.

Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x+a_0$ . Construct the polynomials $p_1(x)=a_n(x-1)^n+a_{n-1}(x-1)^{n-1}+\dots +a_1(x-1)+a_0$ and $p_2(x)=a_n(x-2)^n+a_{n-1}(x-2)^{n-1}+\dots +a_1(x-2)+a_0$.

Then the polynomial $p-(p_1+p_2)$ has infinitely many roots since $p(i)=p_1(i)+p_2(i)$ for $i$ an integer $\ge 1$ because of the Fibonacci recurrence. Hence we have $p_1+p_2=p$ since a non-zero polynomial of degree $n$ can have at most $n$ zeros.

However $p_1+p_2$ has leading coefficient $2a_n$. So they cannot be equal.

What do you think of my proof. Can you find other proofs? Simpler proofs? harder proofs? The less standard the better. I appreciate proofs which might look excessive in the amount of theory they use, although only if they finish quickly.

Thank you very much.

Regards