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I'm working on the following problem:

Prove that there is a unique polynomial $f$ of degree at most $4$ with rational coefficients such that $f(1) = 1, f(2) = 2, f(3) = 4, f(4) = 8, $ and $f(5) = 16$.

I have a proof of a more general fact that there is a unique polynomial of degree at most $n$ that passes through $n+1$ data points:

Suppose that there are at least two polynomials of degree at most $n$ that pass through the $n + 1$ data points $(x_0, y_0), ... , (x_n, y_n)$, $P(x)$ and $Q(x)$. Define $R(x) = P(x) - Q(x)$. Since both $P(x)$ and $Q(x)$ pass through the $n + 1$ data points, $P(x_i) = Q(x_i)$ ($i = 0,...,n$) $\Rightarrow$ $R(x_i) = P(x_i) - Q(x_i) = 0$ ($i = 0,...,n$). $R(x)$ is an $n$th order polynomial with $n + 1$ zeroes $\Rightarrow$ $R(x)$ is the zero polynomial $\Rightarrow$ $P(x) = Q(x)$.

Can I simply rehash this proof for the above problem, simply by having both $P(x)$ and $Q(x)$ be polynomials with rational coefficients and of degree at most $n = 4$ ? Or is there an entirely different proof required here? If so, how can I go about proving the above problem?

Thanks!

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    $\begingroup$ No, you can't, that would be a circular argument (unless the theorem is constructible, i.e. it gives you an algorithm to construct a polynomial passing through the data points). You would start with assuming that such $P$ and $Q$ exist, only to prove that such $P$ and $Q$ exist and $P=Q$. After you exhibit such a polynomial $P$, you can use the theorem to show that it is unique. Have you come across Lagrange polynomials? $\endgroup$ – Goran Malic Nov 10 at 9:29

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